We will learn how to solve different types of problems on parabola.

**1.** Find the vertex, focus, directrix, axis and latusrectum of the parabola y\(^{2}\) - 4x - 4y = 0

**Solution:**

The given equation of the parabola is y\(^{2}\) - 4x - 4y = 0

⇒ y\(^{2}\) - 4y = 4x

⇒ y\(^{2}\) - 4y + 4 = 4x + 4, (Adding 4 on both sides)

⇒ (y - 2)\(^{2}\) = 4(x + 1) ……………………………….. (i)

Shifting the origin to the point (-1, 2) without rotating the
axes and denoting the new coordinates with respect to these axes by X and Y, we
have

x = X + (-1), y = Y + 2 ……………………………….. (ii)

Using these relations equation (i), reduces to

Y\(^{2}\) = 4X……………………………….. (iii)

This is of the form Y\(^{2}\) = 4aX. On comparing, we get 4a = 4 ⇒ a = 1.

The coordinates of the vertex with respect to new axes are (X = 0, Y = 0)

So, coordinates of the vertex with respect to old axes are (-1, 2), [Putting X= 0, Y = 0 in (ii)].

The coordinates of the focus with respect to new axes are (X = 1, Y = 0)

So, coordinates of the focus with respect to old axes are (0, 2), [Putting X= 1, Y = 0 in (ii)].

Equation of the directrix of the parabola with respect to new axes in X = -1

So, equation of the directrix of the parabola with respect to old asex is x = -2, [Putting X = -1, in (ii)].

Equation of the axis of the parabola with respect to new axes is Y = 0.

So, equation of axis with respect to old axes is y = 2, [Putting Y = 0, in (ii)].

The length of the latusrectum is 4 units.

**2.** Find the point on the parabola y\(^{2}\) = 12x at which the
ordinate is double the abscissa.

**Solution: **

The given parabola is y\(^{2}\) = 12x.

Now, let (k, 2k) be the co-ordinates of the required point (k ≠ 0).

Since the point lies (k, 2k) on the parabola y\(^{2}\) = 12x,

Therefore, we get,

(2k)\(^{2}\) = 12k

⇒ 4k\(^{2}\) = 12k

⇒ k = 3 (Since, k ≠ 0, ).

Therefore, the co-ordinates of the required point are (3, 6).

**3.** Write the parametric equation of the parabola (x + 2)\(^{2}\) =
- 4(y + 1).

**Solution:**

The given equation of the parabola is (x + 2)\(^{2}\) = - 4(y + 1).

Then parametric equation of the parabola (x + 2)\(^{2}\) = - 4(y + 1) are

x + 2 = 2t and y + 1 = -t\(^{2}\)

⇒ x = 2t – 2 and y = -t\(^{2}\) – 1.

**4.** Find the equation of the parabola whose co-ordinates of
vertex and focus are (-2, 3) and (1, 3) respectively.

**Solution:**

According to the problem, the ordinates of vertex and focus are equal hence, the axis of the required parabola is parallel to x-axis. Again,

a = abscissa of focus - abscissa of vertex

⇒ a = 1 - (- 2) = 1 + 2 = 3.

Therefore, the equation of the required parabola is

(y - β)\(^{2}\) = 4a (x - α)

⇒ (y - 3)\(^{2}\) = 4 . 3(x + 2)

⇒ y\(^{2}\) - 6y + 9 = 12x + 24

⇒ y\(^{2}\) - 6y - 12x = 15.

**● ****The Parabola**

**Concept of Parabola****Standard Equation of a Parabola****Standard form of Parabola y22 = - 4ax****Standard form of Parabola x22 = 4ay****Standard form of Parabola x22 = -4ay****Parabola whose Vertex at a given Point and Axis is Parallel to x-axis****Parabola whose Vertex at a given Point and Axis is Parallel to y-axis****Position of a Point with respect to a Parabola****Parametric Equations of a Parabola****Parabola Formulae****Problems on Parabola**

**11 and 12 Grade Math**__From Problems on Straight Lines____ to HOME PAGE__

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