Focus of Parabola Coordinates with Examples

Here you will learn how to find the focus of parabola with examples.

Let’s begin –

Focus of Parabola Coordinates

(i) For Parabola \(y^2\) = 4ax :

The coordinates of focus is (a, 0).

(ii) For Parabola \(y^2\) = -4ax :

The coordinates of focus is (-a, 0).

(iii) For Parabola \(x^2\) = 4ay :

The coordinates of focus is (0, a).

(iv) For Parabola \(x^2\) = -4ay :

The coordinates of focus is (0, -a).

(v) For Parabola \((y – k)^2\) = 4a(x – h) :

The coordinates of focus is (h + a, k).

(vi) For Parabola \((x – p)^2\) = 4a(y – q) :

The coordinates of focus is (p, a + q).

Also Read : Different Types of Parabola Equations

Example : For the given parabola, find the coordinates of the foci :

(i) \(y^2\) = 8x

(ii) \(x^2\) = -16y

Solution :

(i) The given parabola is of the form \(y^2\) = 4ax, where 4a = 8 i.e. a = 2.

Hence, the coordinates of the focus are (a, 0) i.e. (2, 0).

(ii) The given parabola is of the form \(x^2\) = -4ay, where 4a = 16 i.e. a = 4.

Hence, the coordinates of the focus are (0, -a) i.e. (0, -4).

Example : Find the coordinates of foci of the parabola \(y^2 – 8y – x + 19\) = 0

Solution : The given equation is

\(y^2 – 8y – x + 19\) = 0  \(\implies\)  \(y^2 – 8y\) = x – 19

\(\implies\)  \(y^2 – 8y + 16\) = x – 19 + 16

\(\implies\)  \((y – 4)^2\) = (x – 3)

The equation is of the form \((y – k)^2\) = 4a(x – h),

On Comparing we get,

4a = 1 i.e. a = 1/4 and k = 4, h = 3

Hence, the coordinates of focus is (h + a, k).

i.e.  ((3 + 1/4), 4) = (13/4, 4).

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