# NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.2

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## Chapter 11 Ex.11.2 Question 1

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for $${y^2} = 12x$$.

### Solution

The given equation is $${y^2} = 12x$$

Here, the coefficient of $$x$$ is positive.

Hence, the parabola opens towards the right.

On comparing this equation with $${y^2} = 4ax$$, we obtain

$4a = 12 \Rightarrow a = 3$

Therefore,

Coordinates of the focus $$F = \left( {a,0} \right) \Rightarrow \left( {3,0} \right)$$

Since the given equation involves $${y^2}$$, the axis of the parabola is the $$x$$-axis.

Equation of directrix, $$x = - a$$, i.e., $$x = - 3$$

Length of latus rectum $$= 4a = 4 \times 3 = 12$$

## Chapter 11 Ex.11.2 Question 2

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for $${x^2} = 6y$$.

### Solution

The given equation is $${x^2} = 6y$$

Here, the coefficient of $$y$$ is positive.

Hence, the parabola opens upwards.

On comparing this equation with $${x^2} = 4ay$$, we obtain

$4a = 6 \Rightarrow a = \frac{3}{2}$

Therefore,

Coordinates of the focus $$F = \left( {0,a} \right) \Rightarrow \left( {0,\frac{3}{2}} \right)$$

Since the given equation involves $${x^2}$$, the axis of the parabola is the $$y$$-axis.

Equation of directrix, $$y = - a$$, i.e., $$y = - \frac{3}{2}$$

Length of latus rectum$$= 4a = 4 \times \frac{3}{2} = 6$$

## Chapter 11 Ex.11.2 Question 3

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for $${y^2} = - 8x$$

### Solution

The given equation is $${y^2} = - 8x$$

Here, the coefficient of $$x$$ is negative.

Hence, the parabola opens towards the left.

On comparing this equation with $${y^2} = - 4ax$$, we obtain

$- 4a = 8 \Rightarrow a = - 2$

Therefore,

Coordinates of the focus $$F = \left( { - a,0} \right) \Rightarrow \left( { - 2,0} \right)$$

Since the given equation involves $${y^2}$$, the axis of the parabola is the $$x$$-axis.

Equation of directrix, $$x = a$$, i.e., $$x = 2$$

Length of latus rectum $$= 4a = 4 \times 2 = 8$$

## Chapter 11 Ex.11.2 Question 4

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for $${x^2} = - 16y$$.

### Solution

The given equation is $${x^2} = - 16y$$

Here, the coefficient of $$y$$ is negative.

Hence, the parabola opens downwards.

On comparing this equation $${x^2} = - 4ay$$, we obtain

$- 4a = - 16 \Rightarrow a = 4$

Therefore,

Coordinates of the focus $$F = \left( {0, - a} \right) = \left( {0, - 4} \right)$$

Since the given equation involves $${x^2}$$, the axis of the parabola is the $$y$$-axis.

Equation of directrix, $$y = a$$ i.e., $$y = 4$$

Length of latus rectum $$4a = 4 \times 4 = 16$$

## Chapter 11 Ex.11.2 Question 5

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for $${y^2} = 10x$$.

### Solution

The given equation is $${y^2} = 10x$$

Here, the coefficient of $$x$$ is positive.

Hence, the parabola opens towards the right.

On comparing this equation with $${y^2} = 4ax$$, we obtain

$4a = 10 \Rightarrow a = \frac{5}{2}$

Therefore,

Coordinates of the focus $$= \left(a,0\right) = \left( {\frac{5}{2},0} \right)$$

Since the given equation involves $${y^2}$$, the axis of the parabola is the $$x$$-axis.

Equation of directrix, $$x = - a$$, i.e., $$x = - \frac{5}{2}$$

Length of latus rectum $$= 4a = 4 \times \frac{5}{2} = 10$$

## Chapter 11 Ex.11.2 Question 6

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for $${x^2} = - 9y$$.

### Solution

The given equation is $${x^2} = - 9y$$

Here, the coefficient of $$y$$ is negative.

Hence, the parabola opens downwards.

On comparing this equation $${x^2} = - 4ay$$, we obtain

$- 4a = - 9 \Rightarrow a = \frac{9}{4}$

Therefore,

Coordinates of the focus$$= \left( {0, - a} \right) = \left( {0, - \frac{9}{4}} \right)$$

Since the given equation involves $${x^2}$$, the axis of the parabola is the $$y - {\rm{axis}}$$.

Equation of directrix, $$y = a$$ , i.e., $$y=\frac{9}{4}$$

Length of latus rectum $$= 4a = 4 \times \frac{9}{4} = 9$$

## Chapter 11 Ex.11.2 Question 7

Find the equation of the parabola that satisfies the following conditions: Focus $$\left( {6,\;0} \right)$$; Directrix $$x = - 6$$.

### Solution

Focus $$\left( {6,\;0} \right)$$; Directrix $$x = - 6$$

Since the focus lies on the $$x$$-axis, the $$x$$-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form $${y^2} = 4ax$$ or $${y^2} = - 4ax$$.

It is also seen that the directrix, $$x = - 6$$ is to the left of the $$y$$-axis while the focus $$\left( {6,\;0} \right)$$ is to right of the $$y$$-axis.

Hence, the parabola is of the form $${y^2} = 4ax$$.

Here, $$a=6$$

Thus, the equation of the parabola is $${{y}^{2}}=24x$$.

## Chapter 11 Ex.11.2 Question 8

Find the equation of the parabola that satisfies the following conditions: Focus $$\left( {0, - 3} \right)$$; Directrix $$y = 3$$

### Solution

Focus $$\left( {0, - 3} \right)$$; Directrix $$y = 3$$

Since the focus lies on the $$y -$$axis, the $$y -$$axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form $${x^2} = 4ay$$  or $${x^2} = - 4ay$$.

It is also seen that the directrix, $$y = 3$$ is above the $$x$$-axis while the focus $$\left( {0, - 3} \right)$$is below the axis.

Hence, the parabola is of the form $${x^2} = - 4ay$$.

Here, $$a = 3$$

Thus, the equation of the parabola is $${x^2} = - 12y$$.

## Chapter 11 Ex.11.2 Question 9

Find the equation of the parabola that satisfies the following conditions: Vertex $$\left( {0,0} \right)$$; Focus $$\left( {3,0} \right)$$

### Solution

Vertex $$\left( {0,0} \right)$$; Focus $$\left( {3,0} \right)$$

Since the vertex of the parabola is $$\left( {0,0} \right)$$and the focus lies on the positive $$x$$-axis, $$x$$-axis is the axis of the parabola, while the equation of the parabola is of the form $${y^2} = 4ax$$

Since the focus is $$\left( 3,0 \right)$$, $$a=3$$

Thus, the equation of the parabola is $${{y}^{2}}=4\times 3\times x$$, i.e., $${{y}^{2}}=12x$$

## Chapter 11 Ex.11.2 Question 10

Find the equation of the parabola that satisfies the following conditions:

Vertex $$\left( {0,0} \right)$$; Focus $$\left( { - 2,0} \right)$$

### Solution

Vertex $$\left( {0,0} \right)$$; Focus $$\left( { - 2,0} \right)$$

Since the vertex of the parabola is $$\left( {0,0} \right)$$and the focus lies on the negative $$x$$-axis, $$x$$-axis is the axis of the parabola, while the equation of the parabola is of the form $${y^2} = - 4ax$$

Since the focus is $$\left( { - 2,0} \right)$$, $$a = 2$$

Thus, the equation of the parabola is $${y^2} = - 4 \times 2 \times x,$$ i.e., $${y^2} = - 8x$$

## Chapter 11 Ex.11.2 Question 11

Find the equation of the parabola that satisfies the following conditions: Vertex $$\left( {0,0} \right)$$passing through $$\left( {2,3} \right)$$ and axis is along $$x$$-axis.

### Solution

Since the vertex is $$\left( {0,0} \right)$$ and the axis of the parabola is the $$x$$-axis, the equation of the parabola is either of the form $${y^2} = 4ax$$ or $${y^2} = - 4ax$$

The parabola passes through point $$\left( {2,3} \right)$$, which lies in the first quadrant.

Therefore, the equation of the parabola is of the form $${y^2} = 4ax$$, while point $$\left( {2,3} \right)$$ must satisfy the equation $${y^2} = 4ax$$

Hence,

${3^2} = 4a \times 2 \Rightarrow a = \frac{9}{8}$

Thus, the equation of the parabola is

\begin{align}&\Rightarrow\; {y^2} = 4 \times \frac{9}{8} \times x\\&\Rightarrow\; {y^2} = \frac{9}{2}x\\&\Rightarrow\; 2{y^2} = 9x\end{align}

## Chapter 11 Ex.11.2 Question 12

Find the equation of the parabola that satisfies the following conditions: Vertex $$\left( {0,0} \right)$$ passing through $$\left( {5,2} \right)$$ and symmetric with respect to $$y -$$axis.

### Solution

Since the vertex is $$\left( {0,0} \right)$$ and the parabola is symmetric about the $$y -$$axis, the equation of the parabola is either of the form $${x^2} = 4ay$$ or $${x^2} = - 4ay$$

The parabola passes through point $$\left( {5,2} \right)$$, which lies in the first quadrant.

Therefore, the equation of the parabola is of the form $${x^2} = 4ay$$, while point $$\left( {5,2} \right)$$ must satisfy the equation $${x^2} = 4ay$$

Hence,

${5^2} = 4a \times 2 \Rightarrow 25 = 8a \Rightarrow a = \frac{{25}}{8}$

Thus, the equation of the parabola is

\begin{align}&\Rightarrow\; {x^2} = 4 \times \frac{{25}}{8} \times y\\&\Rightarrow\;{x^2} = \frac{{25}}{2}y\\&\Rightarrow \;2{x^2} = 25y\end{align}

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