Question

Tangents are drawn to $y^2$ = $4ax$ at the point where the line $lx$ + $my$ + $n$ = 0 meets the parabola . Intersection point of these tangents is

• A. $(\frac{-n}{1},\frac{2am}{1})$
• B. $(\frac{n}{1},\frac{am}{1})$
• C. $(\frac{n}{1},\frac{-2am}{1})$
• D. $(\frac{n}{1},\frac{am}{-1})$

Answer 1

The correct option is C $(\frac{n}{1},\frac{-2am}{1})$

Let the point of intersection of tangents be $(x_1,y_1)$ then

the equation of the chord of contact of tangents drawn from $P$ to the parabola $y^2=4ax$ is

$y{y}_1=2a(x+x_1)$.

Clearly, $lx+my+n=0$ is also the chord of contact of tangents.

Therefore, $y{y}_1=2a(x+x_1)$ and $lx+my+n=0$ represent the same line.

$\therefore \frac{2a}{l}=\frac{-y_1}{m}=\frac{2a{x}_1}{n}$

$\Rightarrow =\frac{n}{l}$ and $y_1=\frac{2am}{l}$

Hence, the required point is $(\frac{n}{l},-\frac{2am}{l})$