# NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.3

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## Chapter 6 Ex.6.3 Question 1

Find the slope of the tangent to the curve $$y = 3{x^4} - 4x$$at $$x = 4$$.

### Solution

The given curve is $$y = 3{x^4} - 4x$$

Then, the slope of the tangent to the given curve at $$x = 4$$ is given by,

\begin{align}{\left. {\frac{{dy}}{{dx}}} \right]_{x = 4}}& = \frac{d}{{dx}}{\left. {\left( {3{x^4} - 4x} \right)} \right]_{x = 4}}\\& = {\left. {12{x^3} - 4} \right]_{x = 4}}\\ &= 12{\left( 4 \right)^3} - 4\\& = 12\left( {64} \right) - 4\\& = 764\end{align}

## Chapter 6 Ex.6.3 Question 2

Find the slope of the tangent to the curve $$y = \frac{{x - 1}}{{x - 2}},x \ne 2$$ at $$x = 10$$.

### Solution

The given curve is $$y = \frac{{x - 1}}{{x - 2}}$$

Therefore,

\begin{align}\frac{{dy}}{{dx}}& = \frac{d}{{dx}}\left( {\frac{{x - 1}}{{x - 2}}} \right)\\& = \frac{{\left( {x - 2} \right)\left( 1 \right) - \left( {x - 1} \right)\left( 1 \right)}}{{{{\left( {x - 2} \right)}^2}}}\\ &= \frac{{x - 2 - x + 1}}{{{{\left( {x - 2} \right)}^2}}}\\ &= \frac{{ - 1}}{{{{\left( {x - 2} \right)}^2}}}\end{align}

Now, the slope of the tangent to the given curve at $$x=10$$ is given by,

\begin{align}{\left. {\frac{{dy}}{{dx}}} \right]_{x = 10}}& = {\left. {\frac{{ - 1}}{{{{\left( {x - 2} \right)}^2}}}} \right]_{x = 10}}\\ &= \frac{{ - 1}}{{{{\left( {10 - 2} \right)}^2}}}\\ &= \frac{{ - 1}}{{64}}\end{align}

## Chapter 6 Ex.6.3 Question 3

Find the slope of the tangent to curve $$y = {x^3} - x + 1$$at the point whose x-coordinate is $$2$$.

### Solution

The given curve is $$y = {x^3} - x + 1$$

Therefore,

\begin{align}\frac{{dy}}{{dx}}& = \frac{d}{{dx}}\left( {{x^3} - x + 1} \right)\\ &= 3{x^2} - 1\end{align}

Now, the slope of the tangent at the point where the $$x-$$coordinate is $$2$$ is given by,

\begin{align}{\left. {\frac{{dy}}{{dx}}} \right]_{x = 2}}& = {\left. {3{x^2} - 1} \right]_{x = 2}}\\ &= 3{\left( 2 \right)^2} - 1\\ &= 12 - 1\\& = 11\end{align}

## Chapter 6 Ex.6.3 Question 4

Find the slope of the tangent to curve $$y = {x^3} - 3x + 2$$at the point whose $$x-$$coordinate is $$3$$.

### Solution

The given curve is $$y = {x^3} - 3x + 2$$

Therefore,

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{x^3} - 3x + 2} \right)\\ &= 3{x^2} - 3\end{align}

Now, the slope of the tangent at the point where the $$x-$$coordinate is $$3$$ is given by,

\begin{align}{\left. {\frac{{dy}}{{dx}}} \right]_{x = 2}} &= {\left. {3{x^2} - 3} \right]_{x = 3}}\\ &= 3{\left( 3 \right)^2} - 3\\ &= 27 - 3\\ &= 24\end{align}

## Chapter 6 Ex.6.3 Question 5

Find the slope of the normal to the curve $$x = a{\cos ^3}\theta ,y = a{\sin ^3}\theta$$ at $$\theta = \frac{\pi }{4}$$.

### Solution

The given curve is $$x = a{\cos ^3}\theta$$ and $$y = a{\sin ^3}\theta$$

Therefore,

\begin{align}\frac{{dx}}{{d\theta }}& = \frac{d}{{d\theta }}\left( {a{{\cos }^3}\theta } \right)\\ &= - 3a{\cos ^2}\theta \sin \theta \\\frac{{dy}}{{d\theta }} &= \frac{d}{{d\theta }}\left( {a{{\sin }^3}\theta } \right)\\ &= 3a{\sin ^2}\theta \left( {\cos \theta } \right)\end{align}

Hence,

\begin{align}\frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}} = \frac{{3a{{\sin }^2}\theta \cos \theta }}{{ - 3a{{\cos }^2}\theta \sin \theta }}\\ &= - \frac{{\sin \theta }}{{\cos \theta }}\\& = - \tan \theta \end{align}

Now, the slope of the tangent to a curve at $$\theta = \frac{\pi }{4}$$ is given by,

\begin{align}{\left. {\frac{{dy}}{{dx}}} \right]_{\theta = \frac{\pi }{4}}} &= - {\left. {\tan \theta } \right]_{\theta = \frac{\pi }{4}}}\\ &= - \tan \frac{\pi }{4}\\& = - 1\end{align}

Hence, the slope of the normal at $$\theta = \frac{\pi }{4}$$ is given by,

$$\frac{{ - 1}}{{{\text{slope of the tangent at }}\theta = \frac{\pi }{4}}} = \frac{{ - 1}}{{ - 1}} = 1$$

## Chapter 6 Ex.6.3 Question 6

Find the slope of the normal to the curve $$x = 1 - a\sin \theta$$ and $$y = b{\cos ^2}\theta$$ at $$\theta = \frac{\pi }{2}$$.

### Solution

It is given that $$x = 1 - a\sin \theta$$ and $$y = b{\cos ^2}\theta$$

Therefore,

\begin{align}\frac{{dx}}{{d\theta }}& = \frac{d}{{d\theta }}\left( {1 - a\sin \theta } \right)\\& = - a\cos \theta \\\frac{{dy}}{{d\theta }} &= \frac{d}{{d\theta }}\left( {b{{\cos }^2}\theta } \right)\\& = - 2b\sin \theta \cos \theta \end{align}

Hence,

\begin{align}\frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}} = \frac{{ - 2b\sin \theta \cos \theta }}{{ - a\cos \theta }}\\ &= \frac{{2b}}{a}\sin \theta \end{align}

Now, the slope of the tangent at $$\theta = \frac{\pi }{2}$$ is given by,

\begin{align}{\left. {\frac{{dy}}{{dx}}} \right]_{\theta = \frac{\pi }{2}}} &= {\left. {\frac{{2b}}{a}\sin \theta } \right]_{\theta = \frac{\pi }{2}}}\\& = \frac{{2b}}{a}\sin \frac{\pi }{2}\\& = \frac{{2b}}{a}\end{align}

Hence, the slope of normal at $$\theta = \frac{\pi }{2}$$ is given by,

$\frac{{ - 1}}{{{\text{slope of the tangent at }}\theta = \frac{\pi }{2}}} = \frac{{ - 1}}{{\left( {\frac{{2b}}{a}} \right)}} = - \frac{a}{{2b}}$

## Chapter 6 Ex.6.3 Question 7

Find the points at which tangent to the curve $$y = {x^3} - 3{x^2} - 9x + 7$$ is parallel to the $$x-$$axis.

### Solution

The given curve is $$y = {x^3} - 3{x^2} - 9x + 7$$

Therefore,

\begin{align}\frac{{dy}}{{dx}}& = \frac{d}{{dx}}\left( {{x^3} - 3{x^2} - 9x + 7} \right)\\ &= 3{x^2} - 6x - 9\end{align}

Since tangent is parallel to the $$x-$$axis if the slope of the tangent is zero.

Hence,

\begin{align}&3{x^2} - 6x - 9 = 0\\ &\Rightarrow {x^2} - 2x - 3 = 0\\& \Rightarrow \left( {x - 3} \right)\left( {x + 1} \right) = 0\\& \Rightarrow x = 3{\rm{ or }}x = - 1\end{align}

When, $$x = 3$$

Then,

\begin{align}y &= {\left( 3 \right)^3} - 3{\left( 3 \right)^2} - 9\left( 3 \right) + 7\\ &= 27 - 27 - 27 + 7\\ &= - 20\end{align}

When, $$x = - 1$$

Then,

\begin{align}y &= {\left( { - 1} \right)^3} - 3{\left( { - 1} \right)^2} - 9\left( { - 1} \right) + 7\\ &= - 1 - 3 + 9 + 7\\ &= 12\end{align}

Thus, the points at which the tangent is parallel to the x-axis are $$\left( {3, - 20} \right)$$ and $$\left( { - 1,12} \right)$$.

## Chapter 6 Ex.6.3 Question 8

Find a point on the curve $$y = {\left( {x - 2} \right)^2}$$ at which the tangent is parallel to the chord joining the points $$\left( {2,0} \right)$$ and $$\left( {4,4} \right)$$.

### Solution

If a tangent is parallel to the chord joining the points $$\left( {2,0} \right)$$ and $$\left( {4,4} \right)$$

Then, the slope of the tangent $$=$$ the slope of the chord.

Hence, the slope of chord is $$\frac{{4 - 0}}{{4 - 2}} = \frac{4}{2} = 2$$

Now, the slope of the tangent to the given curve is,

$$\frac{{dy}}{{dx}} = 2\left( {x - 2} \right)$$

Since, the slope of the tangent = the slope of the chord.

Hence,

\begin{align}&2\left( {x - 2} \right) = 2\\ &\Rightarrow x - 2 = 1\\ &\Rightarrow x = 3\end{align}

When, $$x = 3$$

Then,

\begin{align}&y = {\left( {3 - 2} \right)^2}\\& = 1\end{align}

Hence, the point on the curve is $$\left( {3,1} \right)$$.

## Chapter 6 Ex.6.3 Question 9

Find the point on the curve $$y = {x^3} - 11x + 5$$ at which the tangent is $$y = x - 11$$.

### Solution

The given curve is $$y = {x^3} - 11x + 5$$

The equation of the tangent to the given curve is $$y = x - 11$$

Therefore, slope of the tangent is $$1$$.

Now, the slope of the tangent to the given curve is,

\begin{align}\frac{{dy}}{{dx}}& = \frac{d}{{dx}}\left( {{x^3} - 11x + 5} \right)\\& = 3{x^2} - 11\end{align}

Hence,

\begin{align}&3{x^2} - 11 = 1\\& \Rightarrow 3{x^2} = 12\\& \Rightarrow {x^2} = 4\\& \Rightarrow x = \pm 2\end{align}

When, $$x = 2$$

Then,

\begin{align}y &= {\left( 2 \right)^3} - 11\left( 2 \right) + 5\\& = 8 - 22 + 5\\& = - 9\end{align}

When, $$x = - 2$$

Then,

\begin{align}y &= {\left( { - 2} \right)^3} - 11\left( { - 2} \right) + 5\\& = - 8 + 22 + 5\\& = 19\end{align}

Thus, the points are $$\left( {2, - 9} \right)$$ and $$\left( { - 2,19} \right)$$.

## Chapter 6 Ex.6.3 Question 10

Find the equation of all lines having slope $$- 1$$ that are tangents to the curve $$y = \frac{1}{{x - 1}},x \ne 1$$.

### Solution

The given curve is $$y = \frac{1}{{x - 1}},x \ne 1$$

The slope of the tangents to the given curve is given by,

\begin{align}\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\frac{1}{{x - 1}}} \right)\\ = \frac{{ - 1}}{{{{\left( {x - 1} \right)}^2}}}\end{align}

The slope of the tangent is $$- 1$$

So, we have:

\begin{align}&\frac{{ - 1}}{{{{\left( {x - 1} \right)}^2}}} = - 1\\ &\Rightarrow {\left( {x - 1} \right)^2} = 1\\& \Rightarrow x - 1 = \pm 1\\& \Rightarrow x = 2,0\end{align}

When, $$x = 0, \Rightarrow y = - 1$$ and when $$x = 2, \Rightarrow y = 1$$

Thus, there are two tangents to the given curve having slope $$- 1$$ and passing through the points $$\left( {0, - 1} \right)$$ and $$\left( {2,1} \right)$$.

Hence, the equation of the tangent through $$\left( {0, - 1} \right)$$ is given by,

\begin{align}y - \left( { - 1} \right) = - 1\left( {x - 0} \right)\\ \Rightarrow y + 1 = - x\\ \Rightarrow y + x + 1 = 0\end{align}

The equation of the tangent through $$\left( {2,1} \right)$$ is given by,

\begin{align}y - 1 = - 1\left( {x - 2} \right)\\ \Rightarrow y - 1 = x + 2\\ \Rightarrow y + x - 3 = 0\end{align}

Thus, the equations of the required lines are $$y + x + 1 = 0$$ and $$y + x - 3 = 0$$.

## Chapter 6 Ex.6.3 Question 11

Find the equations of all lines having slope 2 which are tangent to the curve $$y = \frac{1}{{x - 3}},x \ne 3$$.

### Solution

The given curve is $$y = \frac{1}{{x - 3}}$$

The slope of the tangents to the given curve is given by,

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {\frac{1}{{x - 3}}} \right)\\& = \frac{{ - 1}}{{{{\left( {x - 3} \right)}^2}}}\end{align}

The slope of the tangent is $$2$$

So, we have:

\begin{align}&\frac{{ - 1}}{{{{\left( {x - 3} \right)}^2}}} = 2\\ &\Rightarrow 2{\left( {x - 3} \right)^2} = - 1\\ &\Rightarrow {\left( {x - 3} \right)^2} = \frac{{ - 1}}{2}\end{align}

It is not possible since the LHS is positive and RHS is negative

Thus, there is no tangent to the curve of the slope $$2$$.

## Chapter 6 Ex.6.3 Question 12

Find the equations of all lines having slope $$0$$ which are tangent to the curve $$y = \frac{1}{{{x^2} - 2x + 3}}$$.

### Solution

The given curve is $$y = \frac{1}{{{x^2} - 2x + 3}}$$

The slope of the tangents to the given curve is given by,

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {\frac{1}{{{x^2} - 2x + 3}}} \right)\\ &= \frac{{ - \left( {2x - 2} \right)}}{{{{\left( {{x^2} - 2x + 3} \right)}^2}}}\\ &= \frac{{ - 2\left( {x - 1} \right)}}{{{{\left( {{x^2} - 2x + 3} \right)}^2}}}\end{align}

The slope of the tangent is $$0$$

So, we have:

\begin{align}\frac{{ - 2\left( {x - 1} \right)}}{{{{\left( {{x^2} - 2x + 3} \right)}^2}}} &= 0\\ \Rightarrow - 2\left( {x - 1} \right) &= 0\\ \Rightarrow x &= 1\end{align}

When, $$x = 1$$

Then,

\begin{align}y &= \frac{1}{{1 - 2 + 3}}\\ &= \frac{1}{2}\end{align}

Hence, the equation of the tangent through $$\left( {1,\frac{1}{2}} \right)$$ is given by,

\begin{align}&y - \frac{1}{2} = 0\left( {x - 1} \right)\\ &\Rightarrow y - \frac{1}{2} = 0\\ &\Rightarrow y = \frac{1}{2}\end{align}

Thus, the equation of the line is $$y = \frac{1}{2}$$.

## Chapter 6 Ex.6.3 Question 13

Find the points on the curve $$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{{16}} = 1$$ at which the tangents are

(i) parallel to $$x-$$axis

(ii) parallel to $$y-$$axis

### Solution

The equation of the given curve is $$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{{16}} = 1$$

On differentiating both sides with respect to $$x$$, we have:

\begin{align} &\Rightarrow \frac{{2x}}{9} + \frac{{2y}}{{16}}.\frac{{dy}}{{dx}} = 0\\ &\Rightarrow \frac{{dy}}{{dx}} = \frac{{ - 16x}}{{9y}}\end{align}

The tangent is parallel to the x-axis if the slope of the tangent is $$\frac{{ - 16x}}{{9y}} = 0$$, which is possible if $$x = 0$$

Thus,

$$\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$$ for $$x = 0$$

Therefore,

\begin{align} &\Rightarrow \frac{{{y^2}}}{{16}} = 1 \hfill \\& \Rightarrow {y^2} = 16 \hfill \\& \Rightarrow y = \pm 4 \hfill \\ \end{align}

Hence, the points are $$\left( {0,4} \right)$$ and $$\left( {0, - 4} \right)$$.

(ii) The tangent is parallel to the y-axis if the slope of the normal is 0, which gives

\begin{align}&\frac{{ - 1}}{{\left( {\frac{{ - 16x}}{{9y}}} \right)}} = 0\\ &\Rightarrow \frac{{9y}}{{16x}} = 0\\& \Rightarrow y = 0\end{align}

Thus,

$$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{{16}} = 1$$ for $$y = 0$$

Therefore,

\begin{align}& \Rightarrow \frac{{{x^2}}}{9} = 1\\ &\Rightarrow {x^2} = 9\\ &\Rightarrow x = \pm 3\end{align}

Hence, the points are $$\left( {3,0} \right)$$ and $$\left( { - 3,0} \right)$$.

## Chapter 6 Ex.6.3 Question 14

Find the equation of the tangents and normal to the given curves at the indicated points

(i) $$y = {x^4} - 6{x^3} + 13{x^2} - 10x + 5$$ at $$\left( {0,5} \right)$$

(ii) $$y = {x^4} - 6{x^3} + 13{x^2} - 10x + 5$$ at $$\left( {1,3} \right)$$

(iii) $$y = {x^3}$$ at $$\left( {1,1} \right)$$

(iv) $$y = {x^2}$$ at $$\left( {0,0} \right)$$

(v) $$x = \cos t,y = \sin t$$ at $$t = \frac{\pi }{4}$$

### Solution

(i) The equation of the curve is $$y = {x^4} - 6{x^3} + 13{x^2} - 10x + 5$$

On differentiating with respect to $$x$$, we get:

\begin{align}&\frac{{dy}}{{dx}} = 4{x^3} - 18{x^2} + 26x - 10\\&{\left. {\frac{{dy}}{{dx}}} \right]_{\left( {0,5} \right)}} = - 10\end{align}

Thus, the slope of the tangent at $$\left( {0,5} \right)$$ is $$- 10$$.

The equation of the tangent is given as:

\begin{align}&y - 5 = - 10\left( {x - 0} \right)\\& \Rightarrow y - 5 = - 10x\\ &\Rightarrow 10x + y = 5\end{align}

Slope of normal at $$\left( {0,5} \right)$$ is

$\frac{{ - 1}}{{{\text{slope of the tangent at }}\left( {0,5} \right)}} = \frac{{ - 1}}{{ - 10}} = \frac{1}{{10}}$

Therefore, the equation of the normal at $$\left( {0,5} \right)$$ is given as:

\begin{align}&y - 5 = \frac{1}{{10}}\left( {x - 0} \right)\\& \Rightarrow 10y - 50 = x\\& \Rightarrow x - 10y + 50 = 0\end{align}

(ii) The equation of the curve is $$y = {x^4} - 6{x^3} + 13{x^2} - 10x + 5$$ at $$\left( {1,3} \right)$$

On differentiating with respect to $$x$$, we get:

\begin{align}&\frac{{dy}}{{dx}} = 4{x^3} - 18{x^2} + 26x - 10\\&{\left. {\frac{{dy}}{{dx}}} \right]_{\left( {1,3} \right)}} = 4 - 18 + 26 - 10 = 2\end{align}

Thus, the slope of the tangent at $$\left( {1,3} \right)$$ is $$2$$.

The equation of the tangent is given as:

\begin{align}&y - 3 = 2\left( {x - 1} \right)\\& \Rightarrow y - 3 = 2x - 2\\ &\Rightarrow y = 2x + 1\end{align}

Slope of normal at $$\left( {1,3} \right)$$ is

$\frac{{ - 1}}{{{\text{slope of the tangent at }}\left( {1,3} \right)}} = \frac{{ - 1}}{2}$

Therefore, the equation of the normal at $$\left( {1,3} \right)$$ is given as:

\begin{align}&y - 3 = \frac{1}{2}\left( {x - 1} \right)\\ &\Rightarrow 2y - 6 = x + 1\\& \Rightarrow x + 2y - 7 = 0\end{align}

(iii) The equation of the curve is $$y = {x^3}$$ at $$\left( {1,1} \right)$$

On differentiating with respect to $$x$$, we get:

\begin{align}&\frac{{dy}}{{dx}} = 3{x^2}\\&{\left. {\frac{{dy}}{{dx}}} \right]_{\left( {1,1} \right)}} = 3{\left( 1 \right)^2} = 3\end{align}

Thus, the slope of the tangent at $$\left( {1,1} \right)$$ is 3.

The equation of the tangent is given as:

\begin{align}&y - 1 = 3\left( {x - 1} \right)\\& \Rightarrow y = 3x - 2\end{align}

Slope of normal at $$\left( {1,1} \right)$$ is

$\frac{{ - 1}}{{{\text{slope of the tangent at }}\left( {1,1} \right)}} = \frac{{ - 1}}{3}$

Therefore, the equation of the normal at $$\left( {1,1} \right)$$ is given as:

\begin{align}&y - 1 = \frac{{ - 1}}{3}\left( {x - 1} \right)\\ &\Rightarrow 3y - 3 = - x + 1\\&\Rightarrow x + 3y - 4 = 0\end{align}

(iv) The equation of the curve is at $$\left( {0,0} \right)$$

On differentiating with respect to $$x$$, we get:

\begin{align}&\frac{{dy}}{{dx}} = 2x\\&{\left. {\frac{{dy}}{{dx}}} \right]_{\left( {0,0} \right)}} = 0\end{align}

Thus, the slope of the tangent at $$\left( {0,0} \right)$$ is 0.

The equation of the tangent is given as:

\begin{align}&y - 0 = 0\left( {x - 0} \right)\\ &\Rightarrow y = 0\end{align}

Slope of normal at $$\left( {0,0} \right)$$ is

$\frac{{ - 1}}{{{\text{slope of the tangent at }}\left( {0,0} \right)}} = \frac{{ - 1}}{0}$, which is not defined.

Therefore, the equation of the normal at $$\left( {0,0} \right)$$ is given as $$x = 0$$

(v) The equation of the curve is $$x = \cos t$$ and $$y = \sin t$$ at $$t = \frac{\pi }{4}$$

On differentiating with respect to $$t$$, we get:

$$\frac{{dx}}{{dt}} = - \sin t$$ and $$\frac{{dy}}{{dt}} = \cos t$$

Therefore,

$$\frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{\cos t}}{{ - \sin t}} = - \cot t$$

Hence,

$${\left. {\frac{{dy}}{{dx}}} \right]_{t = \frac{\pi }{4}}} = - \cot t = - 1$$

Thus, the slope of the tangent at $$t = \frac{\pi }{4}$$ is $$- 1$$.

Hence,

$$x = \frac{1}{{\sqrt 2 }}$$ and $$y = \frac{1}{{\sqrt 2 }}$$

The equation of the tangent is given as:

\begin{align}& \Rightarrow y - \frac{1}{{\sqrt 2 }} = - 1\left( {x - \frac{1}{{\sqrt 2 }}} \right)\\& \Rightarrow x + y - \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} = 0\\ &\Rightarrow x + y - \sqrt 2 = 0\end{align}

Slope of normal at $$t = \frac{\pi }{4}$$ is

$\frac{{ - 1}}{{{\text{slope of the tangent at }}t = \frac{\pi }{4}}} = \frac{{ - 1}}{{ - 1}} = 1$

Therefore, the equation of the normal at $$t = \frac{\pi }{4}$$ is given as:

\begin{align}&y - \frac{1}{{\sqrt 2 }} = 1\left( {x - \frac{1}{{\sqrt 2 }}} \right)\\ &\Rightarrow x = y\end{align}

## Chapter 6 Ex.6.3 Question 15

Find the equation of the tangent line to the curve $$y = {x^2} - 2x + 7$$ which is

(a) parallel to the line $$2x - y + 9 = 0$$

(b) perpendicular to the line $$5y - 15x = 13$$

### Solution

(a) The equation of the curve is $$y = {x^2} - 2x + 7$$

On differentiating with respect to $$x$$ , we get:

$$\frac{{dy}}{{dx}} = 2x - 2$$

The equation of the line is $$2x - y + 9 = 0$$

$$\Rightarrow y = 2x + 9$$

This is of the form $$y = mx + c$$

Hence, the slope of line is $$2$$

If a tangent is parallel to the line $$2x - y + 9 = 0$$, then the slope of the tangent is equal to the slope of the line.

Therefore, we have:

\begin{align}&2 = 2x - 2\\& \Rightarrow 2x = 4\\ &\Rightarrow x = 2\end{align}

Now, $$x = 2$$

Then,

\begin{align} &\Rightarrow y = 4 - 4 + 7\\& \Rightarrow y = 7\end{align}

Thus, the equation of tangent passing through $$\left( {2,7} \right)$$ is given by,

\begin{align}&y - 7 = 2\left( {x - 2} \right)\\ &\Rightarrow y - 2x - 3 = 0\end{align}

(b) perpendicular to the line $$5y - 15x = 13$$

$$\Rightarrow y = 3x + \frac{{13}}{5}$$

This is of the form $$y = mx + c$$

Hence, the slope of line is 3

If a tangent is perpendicular to the line $$5y - 15x = 13$$ , then the slope of the tangent is,

$$\frac{{ - 1}}{{{\text{slope of the line}}}} = \frac{{ - 1}}{3}$$

Therefore, we have:

\begin{align}&2x - 2 = \frac{{ - 1}}{3}\\ &\Rightarrow 2x = \frac{{ - 1}}{3} + 2\\ &\Rightarrow 2x = \frac{5}{3}\\& \Rightarrow x = \frac{5}{6}\end{align}

Now, $$x = \frac{5}{6}$$

Then,

\begin{align}y& = \frac{{25}}{{36}} + \frac{{10}}{6} + 7\\ &= \frac{{25 - 60 + 252}}{{36}}\\ &= \frac{{217}}{{36}}\end{align}

Thus, the equation of tangent passing through $$\left( {\frac{5}{6},\frac{{217}}{{36}}} \right)$$ is given by,

\begin{align}&y - \frac{{217}}{{36}} = \frac{1}{3}\left( {x - \frac{5}{6}} \right)\\& \Rightarrow \frac{{36y - 217}}{{36}} = \frac{{ - 1}}{{18}}\left( {6x - 5} \right)\\& \Rightarrow 36y - 217 = - 2\left( {6x - 5} \right)\\& \Rightarrow 36y - 217 = - 12x + 10\\ &\Rightarrow 36y + 12x - 227 = 0\end{align}

## Chapter 6 Ex.6.3 Question 16

Show that the tangents to the curve $$y = 7{x^3} + 11$$ at the points where $$x = 2$$ and $$x = - 2$$ are parallel.

### Solution

The equation of the given curve is $$y = 7{x^3} + 11$$.

Therefore,

$\frac{{dy}}{{dx}} = 21{x^2}$

Thus, the slope of the tangent at the point where $$x = 2$$ is given by,

${\left. {\frac{{dy}}{{dx}}} \right]_{x = 2}} = 21{\left( 2 \right)^2} = 84$

Also, the slope of the tangent at the point where $$x = - 2$$ is given by,

${\left. {\frac{{dy}}{{dx}}} \right]_{x = - 2}} = 21{\left( { - 2} \right)^2} = 84$

It is observed clearly that the slopes of the tangents at the points where $$x = 2$$ and $$x = - 2$$ are equal.

Hence, the two tangents are parallel.

## Chapter 6 Ex.6.3 Question 17

Find the points on the curve $$y = {x^3}$$ at which the slope of the tangent is equal to the y-coordinates of the point.

### Solution

The equation of the given curve is $$y = {x^3}$$

Therefore,

$\frac{{dy}}{{dx}} = 3{x^2}$

When the slope of the tangent is equal to the y-coordinate of the point, then According to the question, $$y = 3{x^2}$$

Also, we have $$y = {x^3}$$

Therefore,

\begin{align}&3{x^2} = {x^3}\\& \Rightarrow {x^2}\left( {x - 3} \right) = 0\\& \Rightarrow x = 0,x = 3\end{align}

When, $$x = 0, \Rightarrow y = 3$$ and $$x = 3, \Rightarrow y = 3{\left( 3 \right)^2} = 27$$

Thus, the points are $$\left( {0,0} \right)$$ and $$\left( {3,27} \right)$$.

## Chapter 6 Ex.6.3 Question 18

For the curve $$y = 4{x^3} - 2{x^5}$$, find all the points at which the tangent passes through the origin.

### Solution

The equation of the given curve is $$y = 4{x^3} - 2{x^5}$$

Therefore,

$$\frac{{dy}}{{dx}} = 12{x^2} - 10{x^4}$$

Hence, the slope of the tangent at the point $$\left( {x,y} \right)$$ is $$12{x^2} - 10{x^4}$$

Thus, the equation of the tangent at $$\left( {x,y} \right)$$ is given by,

$$Y - y = \left( {12{x^2} - 10{x^4}} \right)\left( {X - x} \right)$$

When the tangent passes through the origin $$\left( {0,0} \right)$$, $$X = Y = 0$$.

Therefore,

\begin{align} &- y = \left( {12{x^2} - 10{x^4}} \right)\left( { - x} \right)\\&y = 12{x^3} - 10{x^5}\end{align}

Also, we have $$y = 4{x^3} - 2{x^5}$$

Hence,

\begin{align}&\therefore 12{x^3} - 10{x^5} = 4{x^3} - 2{x^5}\\& \Rightarrow 8{x^5} - 8{x^3} = 0\\& \Rightarrow {x^5} - {x^3} = 0\\ &\Rightarrow {x^3}\left( {{x^2} - 1} \right) = 0\\& \Rightarrow x = 0, \pm 1\end{align}

When, $$x = 0, \Rightarrow y = 4{\left( 0 \right)^3} - 2{\left( 0 \right)^5} = 0$$

When, $$x = 1, \Rightarrow y = 4{\left( 1 \right)^3} - 2{\left( 1 \right)^5} = 2$$

When, $$x = - 1, \Rightarrow y = 4{\left( { - 1} \right)^3} - 2{\left( { - 1} \right)^5} = - 2$$

Thus, the points are $$\left( {0,0} \right)$$, $$\left( {1,2} \right)$$ and $$\left( { - 1, - 2} \right)$$

## Chapter 6 Ex.6.3 Question 19

Find the points on the curve $${x^2} + {y^2} - 2x - 3 = 0$$ at which the tangents are parallel to the x-axis.

### Solution

The equation of the given curve is $${x^2} + {y^2} - 2x - 3 = 0$$

On differentiating with respect to $$x$$, we have:

\begin{align}&2x + 2y\frac{{dy}}{{dx}} - 2 = 0\\ &\Rightarrow y\frac{{dy}}{{dx}} = 1 - x\\& \Rightarrow \frac{{dy}}{{dx}} = \frac{{1 - x}}{y}\end{align}

Now, the tangents are parallel to the x-axis if the slope of the tangent is 0.

Therefore,

\begin{align}&\frac{{1 - x}}{y} = 0\\& \Rightarrow 1 - x = 0\\& \Rightarrow x = 1\end{align}

But we have $${x^2} + {y^2} - 2x - 3 = 0$$ for $$x = 1$$

Hence,

\begin{align} &\Rightarrow {y^2} = 4\\& \Rightarrow y = \pm 2\end{align}

Thus, the points are $$\left( {1,2} \right)$$ and $$\left( {1, - 2} \right)$$.

## Chapter 6 Ex.6.3 Question 20

Find the equation of the normal at the point $$\left( {a{m^2},a{m^3}} \right)$$ for the curve $$a{y^2} = {x^3}$$.

### Solution

The equation of the given curve is $$a{y^2} = {x^3}$$

On differentiating with respect to$$x$$, we have:

\begin{align}&2ay\frac{{dy}}{{dx}} = 3{x^2}\\& \Rightarrow \frac{{dy}}{{dx}} = \frac{{3{x^2}}}{{2ay}}\end{align}

The slope of a tangent to the curve at $$\left( {a{m^2},a{m^3}} \right)$$ is

\begin{align}{\left. {\frac{{dy}}{{dx}}} \right]_{\left( {a{m^2},a{m^3}} \right)}}& = \frac{{3{{\left( {a{m^2}} \right)}^2}}}{{2a\left( {a{m^3}} \right)}}\\& = \frac{{3{a^2}{m^4}}}{{2{a^2}{m^3}}}\\& = \frac{{3m}}{2}\end{align}

Therefore, the slope of normal at $$\left( {a{m^2},a{m^3}} \right)$$ is

$\frac{{ - 1}}{{{\text{slope of the tangent at }}\left( {a{m^2},a{m^3}} \right)}} = \frac{{ - 1}}{{\left( {\frac{{3m}}{2}} \right)}} = \frac{{ - 2}}{{3m}}$

Hence, the equation of the normal at $$\left( {a{m^2},a{m^3}} \right)$$ is given by,

\begin{align}&y - a{m^3} = \frac{{ - 2}}{{3m}}\left( {x - a{m^2}} \right)\\& \Rightarrow 3my - 3a{m^4} = - 2x + 2a{m^2}\\& \Rightarrow 2x + 3my - a{m^2}\left( {2 + 3{m^2}} \right) = 0\end{align}

## Chapter 6 Ex.6.3 Question 21

Find the equation of the normal to the curve $$y = {x^3} + 2x + 6$$ which are parallel to the line $$x + 14y + 4 = 0$$.

### Solution

The equation of the given curve is $$y = {x^3} + 2x + 6$$

The slope of the tangent to the given curve at any point $$\left( {x,y} \right)$$ is given by,

$\frac{{dy}}{{dx}} = 3{x^2} + 2$

Therefore, slope of the normal to the given curve is,

$\frac{{ - 1}}{{{\text{slope of the tangent}}}} = \frac{{ - 1}}{{3{x^2} + 2}}$

The equation of the given line is $$x + 14y + 4 = 0$$

$$\Rightarrow y = - \frac{1}{{14}}x - \frac{4}{{14}}$$ , which is the form of $$y = mx + c$$

Hence,

\begin{align}&\frac{{ - 1}}{{3{x^2} + 2}} = \frac{{ - 1}}{{14}}\\ &\Rightarrow 3{x^2} + 2 = 14\\& \Rightarrow 3{x^2} = 12\\& \Rightarrow {x^2} = 4\\& \Rightarrow x = \pm 2\end{align}

When, $$x = 2, \Rightarrow y = 8 + 4 + 6 = 18$$

When, $$x = - 2, \Rightarrow y = - 8 - 4 + 6 = - 6$$

Therefore, there are two normal to the given curve with slope $$\frac{{ - 1}}{{14}}$$ and passing through the points $$\left( {2,18} \right)$$ and $$\left( { - 2, - 6} \right)$$.

Thus, the equation of the normal through $$\left( {2,18} \right)$$ is

\begin{align}&y - 18 = \frac{{ - 1}}{{14}}\left( {x - 2} \right)\\ &\Rightarrow 14y - 252 = x + 2\\& \Rightarrow x + 14y - 254 = 0\end{align}

And the equation of the normal through $$\left( { - 2, - 6} \right)$$ is

\begin{align}&y - \left( { - 6} \right) = \frac{{ - 1}}{{14}}\left( {x - \left( { - 2} \right)} \right)\\& \Rightarrow 14y + 84 = - x - 2\\& \Rightarrow x + 14y + 86 = 0\end{align}

Hence, the equations of the normal to the given curve are $$x + 14y - 254 = 0$$ and $$x + 14y + 86 = 0$$.

## Chapter 6 Ex.6.3 Question 22

Find the equation of the tangent and normal to the parabola $${y^2} = 4ax$$ at the point $$\left( {a{t^2},2at} \right)$$.

### Solution

The equation of the given parabola is $${y^2} = 4ax$$.

On differentiating both sides with respect to $$x$$, we have:

\begin{align}&2y\frac{{dy}}{{dx}} = 4a\\ &\Rightarrow \frac{{dy}}{{dx}} = \frac{{2a}}{y}\end{align}

Therefore, the slope of the tangent at $$\left( {a{t^2},2at} \right)$$ is

$${\left. {\frac{{dy}}{{dx}}} \right]_{\left( {a{t^2},2at} \right)}} = \frac{{2a}}{{2at}} = \frac{1}{t}$$

Hence, the equation of the tangent at $$\left( {a{t^2},2at} \right)$$ is

\begin{align}&y - 2at = \frac{1}{t}\left( {x - a{t^2}} \right)\\& \Rightarrow ty - 2a{t^2} = x - a{t^2}\\& \Rightarrow ty = x + a{t^2}\end{align}

Now, the slope of the normal at $$\left( {a{t^2},2at} \right)$$ is

$\frac{{ - 1}}{{{\text{slope of the tangent at }}\left( {a{t^2},2at} \right)}} = \frac{{ - 1}}{{\left( {\frac{1}{t}} \right)}} = - t$

Thus, the equation of the normal at $$\left( {a{t^2},2at} \right)$$ is

\begin{align}&y - 2at = - t\left( {x - a{t^2}} \right)\\ &\Rightarrow y - 2at = - tx + a{t^3}\\& \Rightarrow y = - tx + 2at + a{t^3}\end{align}

## Chapter 6 Ex.6.3 Question 23

Prove that the curves $$x = {y^2}$$ and $$xy = k$$ cut at right angles if $$8{k^2} = 1$$.

[Hint: Two curves intersect at right angle if the tangents to the curve at the point of intersection are perpendicular to each other.]

### Solution

The equations of the given curves are $$x = {y^2}$$ and $$xy = k$$

Putting $$x = {y^2}$$ in $$xy = k$$

\begin{align}&{y^3} = k\\& \Rightarrow y = {k^{\frac{1}{3}}}\\& \Rightarrow x = {k^{\frac{2}{3}}}\end{align}

Thus, the point of intersection of the given curves is $$\left( {{k^{\frac{2}{3}}},{k^{\frac{1}{3}}}} \right)$$.

Differentiating $$x = {y^2}$$ with respect to $$x$$, we have:

\begin{align}&1 = 2y\frac{{dy}}{{dx}}\\ &\Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{2y}}\end{align}

Therefore, the slope of the tangent to the curve $$x = {y^2}$$ at $$\left( {{k^{\frac{2}{3}}},{k^{\frac{1}{3}}}} \right)$$ is

${\left. {\frac{{dy}}{{dx}}} \right]_{\left( {{k^{\frac{2}{3}}},{k^{\frac{1}{3}}}} \right)}} = \frac{1}{{2{k^{\frac{1}{3}}}}}$

On differentiating $$xy = k$$ with respect to $$x$$, we have:

\begin{align}&x\frac{{dy}}{{dx}} + y = 0\\ &\Rightarrow \frac{{dy}}{{dx}} = \frac{{ - y}}{x}\end{align}

Now, slope of the tangent to the curve $$xy = k$$ at $$\left( {{k^{\frac{2}{3}}},{k^{\frac{1}{3}}}} \right)$$ is

\begin{align}&{\left. {\frac{{dy}}{{dx}}} \right]_{\left( {{k^{\frac{2}{3}}},{k^{\frac{1}{3}}}} \right)}} = {\left. {\frac{{ - y}}{x}} \right]_{\left( {{k^{\frac{2}{3}}},{k^{\frac{1}{3}}}} \right)}}\\& = - \frac{{{k^{\frac{1}{3}}}}}{{{k^{\frac{2}{3}}}}}\\& = \frac{{ - 1}}{{{k^{\frac{1}{3}}}}}\end{align}

We know that two curves intersect at right angles if the tangents to the curves at the point of intersection i.e., at $$\left( {{k^{\frac{2}{3}}},{k^{\frac{1}{3}}}} \right)$$ are perpendicular to each other.

This implies that we should have the product of the tangents as $$- 1$$.

Thus, the given two curves cut at right angles if the product of the slopes of their respective tangents at $$\left( {{k^{\frac{2}{3}}},{k^{\frac{1}{3}}}} \right)$$ is $$- 1$$. i.e.,

\begin{align}&\left( {\frac{1}{{2{k^{\frac{1}{3}}}}}} \right)\left( {\frac{{ - 1}}{{{k^{\frac{1}{3}}}}}} \right) = - 1\\ &\Rightarrow 2{k^{\frac{2}{3}}} = 1\\& \Rightarrow {\left( {2{k^{\frac{2}{3}}}} \right)^3} = {\left( 1 \right)^3}\\& \Rightarrow 8{k^2} = 1\end{align}

Hence, the given curves cut at right angle if $$8{k^2} = 1$$.

## Chapter 6 Ex.6.3 Question 24

Find the equation of the tangent and normal to the hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$at the point $$\left( {{x_0},{y_0}} \right)$$.

### Solution

Differentiating $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$ with respect to $$x$$, we have:

\begin{align}&\frac{{2x}}{{{a^2}}} - \frac{{2y}}{{{b^2}}}\frac{{dy}}{{dx}} = 0\\& \Rightarrow \frac{{2y}}{{{b^2}}}\frac{{dy}}{{dx}} = \frac{{2x}}{{{a^2}}}\\& \Rightarrow \frac{{dy}}{{dx}} = \frac{{{b^2}x}}{{{a^2}y}}\end{align}

Therefore, the slope of the tangent at $$\left( {{x_0},{y_0}} \right)$$ is

$${\left. {\frac{{dy}}{{dx}}} \right]_{\left( {{x_0},{y_0}} \right)}} = \frac{{{b^2}{x_0}}}{{{a^2}{y_0}}}$$

Hence, the equation of tangent at $$\left( {{x_0},{y_0}} \right)$$ is

\begin{align}&y - {y_0} = \frac{{{b^2}{x_0}}}{{{a^2}{y_0}}}\left( {x - {x_0}} \right)\\ &\Rightarrow {a^2}y{y_0} - {a^2}{y_0}^2 = {b^2}x{x_0} - {b^2}{x_0}^2\\ &\Rightarrow {b^2}x{x_0} - {a^2}y{y_0} - {b^2}{x_0}^2 - {a^2}{y_0}^2 = 0\\& \Rightarrow \frac{{x{x_0}}}{{{a^2}}} - \frac{{y{y_0}}}{{{b^2}}} - \left( {\frac{{{x_0}^2}}{{{a^2}}} - \frac{{{y_0}^2}}{{{b^2}}}} \right) = 0\\ &\Rightarrow \frac{{x{x_0}}}{{{a^2}}} - \frac{{y{y_0}}}{{{b^2}}} - 1 = 0\\& \Rightarrow \frac{{x{x_0}}}{{{a^2}}} - \frac{{y{y_0}}}{{{b^2}}} = 1\end{align}

Now, the slope of normal at $$\left( {{x_0},{y_0}} \right)$$ is

$\frac{{ - 1}}{{{\text{slope of the tangent at }}\left( {{x_0},{y_0}} \right)}} = \frac{{ - {a^2}{y_0}}}{{{b^2}{x_0}}}$

Hence, the equation of normal at $$\left( {{x_0},{y_0}} \right)$$ is

\begin{align}& y - {y_0} = \frac{{ - {a^2}{y_0}}}{{{b^2}{x_0}}}\left( {x - {x_0}} \right)\\& \Rightarrow \frac{{y - {y_0}}}{{{a^2}{y_0}}} = \frac{{ - \left( {x - {x_0}} \right)}}{{{b^2}{x_0}}}\\& \Rightarrow \frac{{y - {y_0}}}{{{a^2}{y_0}}} - \frac{{\left( {x - {x_0}} \right)}}{{{b^2}{x_0}}} = 0 \end{align}

## Chapter 6 Ex.6.3 Question 25

Find the equation of the tangent to the curve $$y = \sqrt {3x - 2}$$which is parallel to the line $$4x - 2y + 5 = 0$$.

### Solution

The equation of the given curve is

The slope of the tangent to the given curve at any point $$\left( {x,y} \right)$$ is given by,

$$\frac{{dy}}{{dx}} = \frac{3}{{2\sqrt {3x - 2} }}$$

The equation of the given line is $$4x - 2y + 5 = 0$$.

$$\Rightarrow y = 2x + \frac{5}{2}$$, which is the form of $$y = mx + c$$

Hence, the slope of line is 2.

Now, the tangent to the given curve is parallel to the line $$4x - 2y + 5 = 0$$ if the slope of the tangent is equal to the slope of the line.

Therefore,

\begin{align}&\frac{3}{{2\sqrt {3x - 2} }} = 2\\ &\Rightarrow \sqrt {3x - 2} = \frac{3}{4}\\& \Rightarrow 3x - 2 = \frac{9}{{16}}\\ &\Rightarrow 3x = \frac{9}{{16}} + 2 = \frac{{41}}{{16}}\\& \Rightarrow x = \frac{{41}}{{48}}\end{align}

When, $$x = \frac{{41}}{{48}}$$

Then,

\begin{align}y &= \sqrt {3\left( {\frac{{41}}{{48}}} \right) - 2} \\ &= \sqrt {\frac{{41}}{{16}} - 2} \\ &= \sqrt {\frac{{41 - 32}}{{16}}} \\ &= \sqrt {\frac{9}{{16}}} \\ &= \frac{3}{4}\end{align}

Thus, equation of the tangent passing through the point $$\left( {\frac{{41}}{{48}},\frac{3}{4}} \right)$$ is

\begin{align}&y - \frac{3}{4} = 2\left( {x - \frac{{41}}{{48}}} \right)\\ &\Rightarrow \frac{{4y - 3}}{4} = 2\left( {\frac{{48x - 41}}{{48}}} \right)\\& \Rightarrow 4y - 3 = \left( {\frac{{48x - 41}}{6}} \right)\\ &\Rightarrow 24y - 18 = 48x - 41\\& \Rightarrow 48x - 24y = 23\end{align}

## Chapter 6 Ex.6.3 Question 26

The slope of the normal to the curve $$y = 2{x^2} + 3\sin x$$at $$x = 0$$ is

(A) $$3$$

(B) $$\frac{1}{3}$$

(C) $$- 3$$

(D) $$- \frac{1}{3}$$

### Solution

The equation of the given curve is $$y = 2{x^2} + 3\sin x$$.

Slope of the tangent to the given curve at $$x = 0$$ is given by,

\begin{align}&{\left. {\frac{{dy}}{{dx}}} \right]_{x = 0}} = {\left. {4x + 3\cos x} \right]_{x = 0}}\\ &= 0 + 3\cos 0\\ &= 3\end{align}

Hence, the slope of the normal to the given curve at $$x = 0$$ is

$\frac{{ - 1}}{{{\text{slope of the tangent at }}\left( {x = 0} \right)}} = \frac{{ - 1}}{3}$

Thus, the correct option is D.

## Chapter 6 Ex.6.3 Question 27

The line $$y = x + 1$$is a tangent to the curve $${y^2} = 4x$$ at the point

(A) $$\left( {1,2} \right)$$

(B) $$\left( {2,1} \right)$$

(C) $$\left( {1, - 2} \right)$$

(D) $$\left( { - 1,2} \right)$$

### Solution

The equation of the given curve is $${y^2} = 4x$$ .

Differentiating with respect to $$x$$, we have:

\begin{align}&2y\frac{{dy}}{{dx}} = 4\\& \Rightarrow \frac{{dy}}{{dx}} = \frac{2}{y}\end{align}

The given line is $$y = x + 1$$ which is of the form $$y = mx + c$$.

Hence, slope of the line is $$1$$

The line $$y = x + 1$$ is a tangent to the given curve if the slope of the line is equal to the slope of the tangent.

Also, the line must intersect the curve.

Thus, we must have:

\begin{align}&\frac{2}{y} = 1\\& \Rightarrow y = 2\end{align}

Therefore,

\begin{align}&y = x + 1\\& \Rightarrow x = y - 1\\& \Rightarrow x = 2 - 1\\& \Rightarrow x = 1\end{align}

Hence, the line $$y = x + 1$$ is a tangent to the given curve at the point $$\left( {1,2} \right)$$.

Thus, the correct answer is A.

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