Question

$P(t=2)$ is a point on the parabola
$y^2=40x$. What is the point of the intersection of tangent at P and the directrix of the parabola?

Answer 1

Given, Parabola: $y^2=40x$
on compare with $y^2=4ax$
we get
$\Rightarrow a=10$
The equation of the tangent at $P\left(t\right)\text{is},ty=x+a{t}^2$

Here, $t=2,a=10$
So equation of tangent is

$\Rightarrow 2y=x+40\cdot \cdot \cdot \left(1\right)$

We know that,equation of directrix
Directrix: $x+a=0\cdots ($

Directrix: $x+10=0\cdot \cdot \cdot \left(2\right)$

To find intersection point of tangent and directrix
On solving (1) and (2)

$\Rightarrow 2y=-10+40$

$\Rightarrow y=15$

$\therefore \text{Required point}\equiv (-10,15)$
Hence, the correct answer is Option b.