**NCERT Solutions for Class 11 Maths Chapter 10 ****Conic Sections**** Ex 10.2**

**In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.**

**Ex 10.2 Class 11 Maths Question 1.**

y^{2 }= 12x

**Solution:**

The given equation of parabola is:

y^{2} = 12x, which is of the form y^{2} = 4ax.

∴ 4a = 12 ⇒ a = 3

∴ Coordinates of the focus are (3, 0).

Axis of parabola is y = 0.

Equation of the directrix is x = -3 ⇒ x + 3 = 0

Length of the latus rectum = 4 × 3 = 12.

**Ex 10.2 Class 11 Maths Question 2.**

x^{2} = 6y

**Solution:**

The given equation of parabola is:

x^{2} = 6y which is of the form x^{2} = 4ay.

**Ex 10.2 Class 11 Maths Question 3.**

y^{2} = –8x

**Solution:**

The given equation of parabola is:

y^{2} = -8x, which is of the form y^{2} = –4ax.

∴ 4a = 8 ⇒ a = 2

∴ Coordinates of the focus are (-2, 0).

Axis of parabola is y = 0.

Equation of the directrix is x = 2 ⇒ x – 2 = 0

Length of the latus rectum = 4 × 2 = 8.

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**Ex 10.2 Class 11 Maths Question 4.**

x^{2} = -16y

**Solution:**

The given equation of parabola is:

x^{2} = -16y, which is of the form x^{2} = -4ay.

∴ 4a = 16 ⇒ a = 4

∴ Coordinates of the focus are (0, -4).

Axis of parabola is x = 0.

Equation of the directrix is y = 4 ⇒ y – 4 = 0

Length of the latus rectum = 4 × 4 = 16.

**Ex 10.2 Class 11 Maths Question 5.**

y^{2 }= 10x

**Solution:**

The given equation of parabola is:

y^{2} = 10x, which is of the form y^{2} = 4ax.

**Ex 10.2 Class 11 Maths Question 6.**

x^{2} = -9y

**Solution:**

The given equation of parabola is:

x^{2} = -9y, which is of the form x^{2} = -4ay.

**In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:**

**Ex 10.2 Class 11 Maths Question 7.**

Focus (6, 0); directrix x = -6

**Solution:**

It is given that the focus (6, 0) lies on the x-axis, therefore, the x-axis is the axis of parabola. Also, the directrix is x = -6, i.e., x = -a and focus (6, 0), i.e., (a, 0). The equation of parabola is of the form y^{2} = 4ax.

Therefore, the required equation of the parabola is:

y^{2} = 4 × 6x ⇒ y^{2} = 24x

**Ex 10.2 Class 11 Maths Question 8.**

Focus (0, -3); directrix y = 3

**Solution:**

It is given that the focus (0, -3) lies on the y-axis, therefore, y-axis is the axis of parabola. Also, the directrix is y = 3, i.e., y = a and focus (0, -3), i.e., (0, -a). The equation of parabola is of the form x^{2} = -4ay.

Therefore, the required equation of the parabola is:

x^{2} = – 4 × 3y ⇒ x^{2} = -12y

**Ex 10.2 Class 11 Maths Question 9.**

Vertex (0, 0); focus (3, 0)

**Solution:**

Since the vertex of the parabola is at (0, 0) and focus is at (3, 0).

∴ y = 0 ⇒ The axis of parabola is along x-axis.

∴ The equation of the parabola is of the form y^{2} = 4ax.

The required equation of the parabola is:

y^{2} = 4 × 3x ⇒ y^{2} = 12x.

**Ex 10.2 Class 11 Maths Question 10.**

Vertex (0, 0); focus (-2, 0)

**Solution:**

Since the vertex of the parabola is at (0, 0) and focus is at (-2, 0).

∴ y = 0 ⇒ The axis of parabola is along x-axis.

∴ The equation of the parabola is of the form y^{2} = –4ax.

Therefore, the required equation of the parabola is:

y^{2} = –4 × 2x ⇒ y^{2} = -8x

**Ex 10.2 Class 11 Maths Question 11.**

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

**Solution:**

Since the vertex of the parabola is at (0, 0) and the axis is along x-axis.

∴ The equation of the parabola is of the form y^{2} = 4ax.

Since the parabola passes through the point (2, 3).

**Ex 10.2 Class 11 Maths Question 12.**

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

**Solution:**

Since the vertex of the parabola is at (0, 0) and it is symmetrical about the y-axis.

∴ The equation of the parabola is of the form x^{2} = 4ay.

Since the parabola passes through the point (5, 2).

**Related Links:**

**NCERT Solutions for Maths Class 9**