Rbse Solutions for Class 11 maths Chapter 10 Exercise 10.2 | Parabola

Get detailed, step-by-step solutions for NCERT Class 11 Maths Chapter 10 Exercise 10.2 . Learn to find the focus, axis, directrix, and length of the latus rectum for parabolas centered at the origin (Q.1-6). Determine the equation of a parabola given its focus and directrix, or its vertex, focus, and a point it passes through (Q.7-12). Covers all four standard forms: $y^2 = \pm 4ax$ and $x^2 = \pm 4ay$.

This exercise focuses on identifying the key features of parabolas from their equations and finding the equation given specific conditions. All parabolas here have the vertex at the origin $(0, 0)$.

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Finding Focus, Axis, Directrix, and Latus Rectum (Exercises 1–6)

The standard forms are $y^2 = 4ax$, $y^2 = -4ax$, $x^2 = 4ay$, and $x^2 = -4ay$.

| Q. No. | Equation | Standard Form | $4a$ value | $a$ value | Axis | Focus $(F)$ | Directrix | Latus Rectum ($|4a|$) |

| :— | :— | :— | :— | :— | :— | :— | :— | :— |

| 1. | $y^2 = 12x$ | $y^2 = 4ax$ | $4a = 12$ | $a = 3$ | $x$-axis | $(a, 0) = \mathbf{(3, 0)}$ | $x = -a \implies \mathbf{x = -3}$ | $\mathbf{12}$ |

| 2. | $x^2 = 6y$ | $x^2 = 4ay$ | $4a = 6$ | $a = 6/4 = 3/2$ | $y$-axis | $(0, a) = \mathbf{(0, 3/2)}$ | $y = -a \implies \mathbf{y = -3/2}$ | $\mathbf{6}$ |

| 3. | $y^2 = -8x$ | $y^2 = -4ax$ | $-4a = -8$ | $a = 2$ | $x$-axis | $(-a, 0) = \mathbf{(-2, 0)}$ | $x = a \implies \mathbf{x = 2}$ | $|-8| = \mathbf{8}$ |

| 4. | $x^2 = -16y$ | $x^2 = -4ay$ | $-4a = -16$ | $a = 4$ | $y$-axis | $(0, -a) = \mathbf{(0, -4)}$ | $y = a \implies \mathbf{y = 4}$ | $|-16| = \mathbf{16}$ |

| 5. | $y^2 = 10x$ | $y^2 = 4ax$ | $4a = 10$ | $a = 10/4 = 5/2$ | $x$-axis | $(a, 0) = \mathbf{(5/2, 0)}$ | $x = -a \implies \mathbf{x = -5/2}$ | $\mathbf{10}$ |

| 6. | $x^2 = -9y$ | $x^2 = -4ay$ | $-4a = -9$ | $a = 9/4$ | $y$-axis | $(0, -a) = \mathbf{(0, -9/4)}$ | $y = a \implies \mathbf{y = 9/4}$ | $|-9| = \mathbf{9}$ |

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Finding the Equation of the Parabola (Exercises 7–12)

7. Focus $(6, 0)$; directrix $x = -6$.

1. The focus $(6, 0)$ is of the form $(a, 0)$, and the directrix $x = -6$ is of the form $x = -a$.
2. This is a rightward opening parabola of the form $y^2 = 4ax$.
3. From the focus, $a = 6$.
4. The equation is $y^2 = 4(6)x$.$$\mathbf{y^2 = 24x}$$

8. Focus $(0, -3)$; directrix $y = 3$.

1. The focus $(0, -3)$ is of the form $(0, -a)$, and the directrix $y = 3$ is of the form $y = a$.
2. This is a downward opening parabola of the form $x^2 = -4ay$.
3. From the focus, $a = 3$.
4. The equation is $x^2 = -4(3)y$.$$\mathbf{x^2 = -12y}$$

9. Vertex $(0, 0)$; focus $(3, 0)$.

1. The vertex is at the origin and the focus is $(3, 0)$, which is of the form $(a, 0)$.
2. This is a rightward opening parabola of the form $y^2 = 4ax$.
3. $a = 3$.
4. The equation is $y^2 = 4(3)x$.$$\mathbf{y^2 = 12x}$$

10. Vertex $(0, 0)$; focus $(-2, 0)$.

1. The vertex is at the origin and the focus is $(-2, 0)$, which is of the form $(-a, 0)$.
2. This is a leftward opening parabola of the form $y^2 = -4ax$.
3. $a = 2$.
4. The equation is $y^2 = -4(2)x$.$$\mathbf{y^2 = -8x}$$

11. Vertex $(0, 0)$, passing through $(2, 3)$, and axis is along the $x$-axis.

1. Since the axis is the $x$-axis, the equation is either $y^2 = 4ax$ (rightward) or $y^2 = -4ax$ (leftward).
2. The parabola passes through $(2, 3)$, which is in the first quadrant ($x>0, y>0$). A leftward parabola ($y^2 = -4ax$) never enters the first quadrant.
3. Therefore, the form is $y^2 = 4ax$.
4. Substitute the point $(2, 3)$:$$3^2 = 4a(2)$$$$9 = 8a \implies a = \frac{9}{8}$$
5. The equation is $y^2 = 4\left(\frac{9}{8}\right)x = \frac{9}{2}x$.$$\mathbf{2y^2 = 9x}$$

12. Vertex $(0, 0)$, passing through $(5, 2)$, and symmetric with respect to the $y$-axis.

1. Symmetry with respect to the $y$-axis means the equation is either $x^2 = 4ay$ (upward) or $x^2 = -4ay$ (downward).
2. The parabola passes through $(5, 2)$, which is in the first quadrant ($x>0, y>0$). A downward parabola ($x^2 = -4ay$) never enters the first quadrant.
3. Therefore, the form is $x^2 = 4ay$.
4. Substitute the point $(5, 2)$:$$5^2 = 4a(2)$$$$25 = 8a \implies a = \frac{25}{8}$$
5. The equation is $x^2 = 4\left(\frac{25}{8}\right)y = \frac{25}{2}y$.$$\mathbf{2x^2 = 25y}$$