*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.*

**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 focus are (3, 0)

Axis of parabola is y = 0

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

Length of latus rectum = 4 x 3 = 12.

**Question 2.**

**x ^{2} = 6y**

**Solution:**

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

**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 focus are (-2, 0)

Axis of parabola is y = 0

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

Length of latus rectum = 4 x 2 = 8.

**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 focus are (0, -4)

Axis of parabola is x = 0

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

Length of latus rectum = 4 x 4 = 16.

**Question 5.**

**y ^{2}= 10x**

**Solution:**

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

**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:*

**Question 7.**

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

**Solution:**

We are given that the focus (6, 0) lies on the x-axis, therefore 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.

The required equation of parabola is

y^{2} = 4 x 6x ⇒ y^{2} = 24x.

**Question 8.**

**Focus (0, -3); directri xy=3**

**Solution:**

We are 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.

The required equation of parabola is

x^{2} = – 4 x 3y ⇒ x^{2} = -12y.

**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 x 3x ⇒ y^{2} = 12x.

**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

The required equation of the parabola is

y^{2} = – 4 x 2x ⇒ y^{2} = -8x.

**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 point (2, 3)

**Question 12.**

**Vertex (0, 0), passing through (5, 2) and symmetric with respect toy-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 point (5, 2)