Question

Find the locus of point of intersection of tangents to the parabola $y^2=4ax$ which include an angle $\alpha$

• A. ${(x-a)}^2{\tan }^2\alpha =y^2-4ax$
• B. ${(x+a)}^2{\sec }^2\alpha =y^2-4ax$
• C. ${(x-a)}^2{\sec }^2\alpha =y^2-4ax$
• D. ${(x+a)}^2{\tan }^2\alpha =y^2-4ax$

Answer 1

The correct option is D ${(x+a)}^2{\tan }^2\alpha =y^2-4ax$

Let the tangents be drawn at the points $t_1$ and

$t_2$
and $(h,k)$ be their point of intersection.

$\therefore h=a{t}_1{t}_2,k=a(t_1+t_2)$ ...(1)

Also, their equations are $t_{1}y=x+a{t}_1^2,t_{2}y=x+a{t}_2^2$
Also their slope are $\frac{1}{t_1}$ and $\frac{1}{t_2}$ ...(2)
If they include an angle $\alpha ,$ then

$\tan \alpha =\frac{(t_2-t_1)}{t_1{t}_2+1}$
or ${\tan }^2\alpha {(1+t_1{t}_2)}^2=\{{(t_1+t_2)}^2-4t_1{t}_2\}$
or ${\tan }^2\alpha {(1+\frac{h}{a})}^2=\{\frac{k^2}{a^2}-\frac{4h}{a}\}$
or ${\tan }^2\alpha {(h+a)}^2=(k^2-4ah)$

Hence, the required locus is${(x+a)}^2{\tan }^2\alpha =y^2-4ax.$