Question

The locus of the point of intersection of the tangents to the parabola $y^2=4ax$ which makes angles ${\theta }_1$ and ${\theta }_2$ with its axis so that $\cot {\theta }_1+\cot {\theta }_2=k$ is

• A. $kx-y=0$
• B. $kx-a=0$
• C. $y-ka=0$
• D. $x-ka=0$

Answer 1

The correct option is C $y-ka=0$

The equation of the parabola is $y^2=4ax$
For a point $(a{t}^2,2at),\tan \theta =\frac{1}{t}$
Given: $\cot {\theta }_1+\cot {\theta }_2=k$
$\therefore t_1+t_2=k$
Equation of tangent is $y=mx+\frac{a}{m},\text{Here}m=\frac{1}{t}$
Let point of intersection be $(\alpha ,\beta )$
Since, $(\alpha ,\beta )$ lies on $y=mx+\frac{a}{m}$
$\Rightarrow \beta =\frac{1}{t}\alpha +at$
$\Rightarrow a{t}^2-\beta t+\alpha =0$
$\therefore a{t}^2-yt+x=0$
Sum of roots
$=t_1+t_2=\frac{y}{a}k=\frac{y}{a}\Rightarrow y=ak$