**NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2**

NCERT Solutions for Class
11 Maths Chapter 11 Conic
Sections Ex 11.2 are the part of NCERT Solutions for Class 11 Maths. Here you
can find the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex
11.2.

**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 11.2 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 11.2 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 11.2 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.

**Ex 11.2 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 11.2 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 11.2 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 11.2 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 11.2 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 11.2 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 11.2 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 11.2 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 11.2 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).

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