Learn Exercise 11.2 with Free Lessons & Tips

The given equation is x² = 6y.

Here, the coefficient of y is positive. Hence, the parabola opens upwards.

On comparing this equation with x² = 4ay, we obtain

∴Coordinates of the focus = (0, a) =

Since the given equation involves x², the axis of the parabola is the y-axis.

Equation of directrix,

Length of latus rectum = 4a = 6

The given equation is y² = –8x.

Here, the coefficient of x is negative. Hence, the parabola opens towards the left.

On comparing this equation with y² = –4ax, we obtain

–4a = –8 ⇒ a = 2

∴Coordinates of the focus = (–a, 0) = (–2, 0)

Since the given equation involves y², the axis of the parabola is the x-axis.

Equation of directrix, x = a i.e., x = 2

Length of latus rectum = 4a = 8

The given equation is x² = –16y.

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

On comparing this equation with x² = – 4ay, we obtain

–4a = –16 ⇒ a = 4

∴Coordinates of the focus = (0, –a) = (0, –4)

Since the given equation involves x², the axis of the parabola is the y-axis.

Equation of directrix, y = a i.e., y = 4

Length of latus rectum = 4a = 16

The given equation is y² = 10x.

Here, the coefficient of x is positive. Hence, the parabola opens towards the right.

On comparing this equation with y² = 4ax, we obtain

∴Coordinates of the focus = (a, 0)

Since the given equation involves y², the axis of the parabola is the x-axis.

Equation of directrix,

Length of latus rectum = 4a = 10

The given equation is x² = –9y.

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

On comparing this equation with x² = –4ay, we obtain

∴Coordinates of the focus =

Since the given equation involves x², the axis of the parabola is the y-axis.

Equation of directrix,

Length of latus rectum = 4a = 9

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 y² = 4ax or

y² = – 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 y² = 4ax.

Here, a = 6

Thus, the equation of the parabola is y² = 24x.

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 x² = 4ay or

x² = – 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 x² = –4ay.

Here, a = 3

Thus, the equation of the parabola is x² = –12y.

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 y² = 4ax.

Since the focus is (3, 0), a = 3.

Thus, the equation of the parabola is y² = 4 × 3 × x, i.e., y² = 12x

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 y² = –4ax.

Since the focus is (–2, 0), a = 2.

Thus, the equation of the parabola is y² = –4(2)x, i.e., y² = –8x

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 y² = 4ax or y² = –4ax.

The parabola passes through point (2, 3), which lies in the first quadrant.

Therefore, the equation of the parabola is of the form y² = 4ax, while point

(2, 3) must satisfy the equation y² = 4ax.

Thus, the equation of the parabola is