NCERT Solutions For Class 10 Maths Chapter 10 Conic Sections Exercise 10.2 - 2025-26

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

Ans: 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’ll get

$4a=12\Rightarrow a=3$

$\therefore $ Coordinates of the focus = $=(a,0)=(3,0)$

Since the given equation involves ${{y}^{2}}$, the axis of the parabola is the x-axis. Equation of directrix, $x=-a\text{  }i.e.,\text{  }x=-3\text{  }i.e.,\text{  }x+3=0$ 

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

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

Ans: 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 

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

$\therefore $Coordinates of the focus $=(0,a)=\left( 0,\dfrac{3}{2} \right)$

Since the given equation involves ${{x}^{2}}$, the axis of the parabola is the y-axis. Equation of directrix, $y=-a$  i.e., $y=\dfrac{-3}{2}$

Length of latus rectum $=4a=6$

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

Ans: 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’ll get

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

$\therefore $ Coordinates of the focus $=(-a,0)=(-2,0)$

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 $=4a=8$

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

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

Here, the coefficient of $y$ is negative. Hence, the parabola opens downwards. 

On comparing this equation with ${{x}^{2}}=-4ay$, we’ll get

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

$\therefore $ Coordinates of the focus $=(0,-a)=(0,-4)$

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 $=4a=16$

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

Ans: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’ll get

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

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

Since the given equation involves ${{y}^{2}}$, the axis of the parabola is the x-axis. Equation of directrix, $x=-a$  i.e., $x=-\dfrac{5}{2}$

Length of latus rectum $=4a=10$

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

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

Here, the coefficient of $y$ is negative. Hence, the parabola opens downwards. 

On comparing this equation with ${{x}^{2}}=-4ay$, we’ll get

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

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

Since the given equation involves ${{x}^{2}}$, the axis of the parabola is the y-axis. Equation of directrix, $y=a$  i.e., $y=\dfrac{9}{4}$

Length of latus rectum $=4a=9$

7. Find the equation of the parabola that satisfies the following conditions: Focus $(6,0);$ directrix $x=-6$

Ans: Focus $(6,0);$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 $(6,0)$ is to the 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 ${{y}^{2}}=24x$.

8. Find the equation of the parabola that satisfies the following conditions: Focus $(0,-3);$ directrix $y=3$

Ans: Focus $=(0,-3);$ 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 $(0,-3)$ is below the x-axis. 

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

Here, $a=3$

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

9. Find the equation of the parabola that satisfies the following conditions: Vertex $(0,0);$ focus $(3,0)$ 

Ans: Vertex $(0,0);$ focus $(3,0)$ 

Since the vertex of the parabola is $(0,0)$ 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 $(3,0)$, $a=3$. 

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

10. Find the equation of the parabola that satisfies the following conditions: Vertex $(0,0)$ focus $(-2,0)$

Ans: Solution 10: Vertex $(0,0)$ focus $(-2,0)$ 

Since the vertex of the parabola is $(0,0)$ 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 $(-2,0),$$a=2.$ 

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

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

Ans: Since the vertex is (0, 0) 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 $(2,3)$, which lies in the first quadrant. Therefore, the equation of the parabola is of the form ${{y}^{2}}=4ax$, while point $(2,3)$ must satisfy the equation ${{y}^{2}}=4ax$.

$\therefore {{(3)}^{2}}=4a(2)\Rightarrow a=\dfrac{9}{8}$

Thus, the equation of the parabola is ${{y}^{2}}=4\left( \dfrac{9}{8} \right)x$

$\begin{align}   & \Rightarrow {{y}^{2}}=\dfrac{9}{2}x \\  & \Rightarrow 2{{y}^{2}}=9x \\ \end{align}$

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

Ans: Since the vertex is $(0,0)$ 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 $(5,2)$, which lies in the first quadrant. Therefore, the equation of the parabola is of the form ${{x}^{2}}=4ay$, while point $(5,2)$ must satisfy the equation ${{x}^{2}}=4ay$. 

$\begin{align}   & \therefore {{(5)}^{2}}=4\times a\times 2 \\  & \Rightarrow 25=8a \\  & \Rightarrow a=\dfrac{25}{8} \\ \end{align}$

Thus, the equation of the parabola is

$\Rightarrow {{x}^{2}}=4\left( \dfrac{25}{8} \right)y$

$\Rightarrow 2{{x}^{2}}=25y$

Class 11 Maths Chapter 10: Exercises Breakdown

Conclusion

NCERT Solutions for Class 11 Maths Chapter 10 - Conic Sections: Exercise 10.2 provide a detailed understanding of parabolas and their properties, including the focus, directrix, and axis of symmetry. These step-by-step solutions help simplify the complex equations of parabolas, making it easier for students to grasp the concepts effectively. By working through these solutions, students can build a solid foundation for more advanced geometry and calculus topics, and gain confidence in tackling exam questions related to conic sections. Make sure to download the FREE PDF for easy access and effective preparation.

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Chapter-Specific NCERT Solutions for Class 11 Maths

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