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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