Parabola

Given


.......... (1)
Equation (2) becomes [using equation (1)]


squaring both sides, we get

where
This equation represents a parabola

Equation of ellipse , where a > b
Given,
But

equation of directrix

The equation of the parabola is
y + 4x + 3 = 0
or
The directrix of the parabola

is X = a.
On comparing the equation and , we
get and
or and
Hence the directrix of the parabola is

On comparing the equation of the parabola
y = 12 x with the standard equation,
y = 4 ax, we get 4 a = 12 or a = 3.
Hence, the focus, point C will be at (3, 0) and the extremities of the latus-rectum AB will be at (a, 2a) and (a, -2a). So the coordinates of A and B are (3, 6) and (3, -6) respectively. Now we need to find the length ( curve AOB) of the parabola. As it is not a straight line so we cannot directly find the length of this curve as we cannot directly apply Pythagorous theorem. Let us consider a small length ds on the parabola. Using pythagorous theorem for this length,


From \left(1\right)\) ,


Using

We get s

Equation of the normal at point on parabola is

It also passes through
So,

Any tangent to parabola is ......(i) It touches the circle . if the length of perpendicular from the centre is equal to radius Hence, the required tangents are and .

Since the parabola passes through the point ,

Equation of tangent to the parabola at
or
or