Question

Find the locus of point of intersection of tangents to the parabola $y^2=4ax$ which include an angle of ${45}^{\circ }.$

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

Answer 1

Answer: (D)

$-{(x+a)}^2=y^2-4ax$

Let the tangents be drawn at the points ${}^{\prime }{t}_1^{\prime }$ and ${}^{\prime }{t}_2^{\prime }$
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
$1/t_1$ and $1/t_2$ ...(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
${(x+a)}^2{\tan }^2\alpha =y^2-4ax.$
In case the tangents include an angle of ${45}^{\circ }$ then
$\tan {45}^{\circ }=1,$ we get the locus as
$-{(x+a)}^2=y^2-4ax$