**Some examples on parabola are given below.**

**Example 1:** Find the vertex, equation of the axis, coordinate of the focus, equation of the directrix of the focus, equation of the directrix and coordinates of the end of the lactus rectum of (y-3)$^{2}$ = 6(x-2)

**Solution:** Comparing with $(y-k)^{2}$ = 4a(x-h)

h = 2, k = 3

Therefore Vertex is (h,k) = (2,3)

Equation of the axis is y = k

i.e., y = 3

4a = 6

a = $\frac{3}{2}$

Focus is (a+h, k) = ($\frac{3}{2}$+ 2, 3)

= ($\frac{7}{2}$,3)

Equation of the directrix is x = -a+h

= -$\frac{3}{2}$+ 2

= $\frac{1}{2}$

i.e., x = $\frac{1}{2}$

Ends of the latus rectum are (a+h, 2a+k)

= ($\frac{3}{2}$ + 2,2.$\frac{3}{2}$ + 3)

= ($\frac{7}{2}$,6)

and (a+h, -2a+k) = ($\frac{3}{2}$ + 2, -2.$\frac{3}{2}$+3)

=($\frac{7}{2}$,0)

**Example 2:** Find the vertex, equation of the axis, focus, equation of the directrix and coordinates of the end of the latus rectum of $(x-1)^{2}$ = -8(y-2)

**Solution:** Comparing (x-1)$^{2}$ = -8(y-2) with (x-h)$^{2}$ = -4a(y-k)

h=1, k = 2

Therefore Vertex is (h,k) = (1,2) and 4a = 8

$\therefore$ a = 2

Equation of the axis is x = h

i.e., x = 1

Focus is (h, -a+k)

= (1,-2+2)

=(1,0)

Equation of the directrix is y =a + k

= 2 + 2

i.e., y = 4

Ends of the latus rectum are

(2a+h, -a+k) and (-2a+h, -a+k)

= (2.2+1,-2+2) and (-2.2+1, -2+2)

=(5,0) and (-3,0)

**Example 3:** Find the equation of the parabola with latus rectum joining the points (-3,1) and (1,1)

**Solution:** Let L = (-3,1) and L' = (1,1)

The focus S is the midpoint of the latus rectum LL'.

$\therefore$ S = (-1,1)

Since LL' is parallel to x axis, the axis of the parabola is parallel to y axis.

The equation of the parabola is of the form

(x-h)$^{2}$= $\pm$4a(y-k)

......1

Length of the latus rectum = 4a

= $\sqrt{(-3-1)^{2}+ (1-1)^{2}}$

= 4

$\therefore$ a = 1

__Case 1:__ If the parabola open upwards, then

S is (h,a+k)= (-1,1)

$\therefore$ h = -1 and a+ k = 1 since a= 1, k = 0

$\therefore$ (h,k) = (-1,0) is the vertex.

$\therefore$ Equation of the parabola is

(x+1)$^{2}$ = 4(1)(y-0)

i.e., (x+1)$^{2}$ = 4y

__Case 2:__ If the parabola opens downwards, then

S is (h,-a+k) = (-1,1)

$\therefore$ h = -1, -a+k = 1 $\therefore$ k = 2

$\therefore$ Equation of the parabola is (x+1)$^{2}$ = -4(y-2)