Question

Two tangents to the parabola $y^2=4ax$ make angle ${\alpha }_1$ and ${\alpha }_2$ with the x-axis. The locus of their point of intersection if $\frac{\cot {\alpha }_1}{\cot {\alpha }_2}=2$

• A. $2y^2=9ax$
• B. $4y^2=9ax$
• C. $y^2=9ax$
• D. None of these

Answer 1

The correct option is A $2y^2=9ax$

$\left(a{t}_1^2,2a{t}_1\right)$

$y=\frac{x}{t_1}+a{t}_1\left[a{t}_2^2,2a{t}_2\right]$

$y=\frac{x}{t^2}+a{t}_2$

Now,

$\frac{\cot {\alpha }_1}{\cot {\alpha }_2}=2\frac{\tan {\alpha }_1}{\tan {\alpha }_2}=2$

$\tan {\alpha }_1=\frac{1}{t_1}\frac{t_1}{t_2}=2\tan {\alpha }_2=\frac{1}{t_2}{t}_1=2t_2$

Then,

$\left(a{t}_1{t}_2,a\left(t_2+t_2\right)\right)\left(2a{t}_2^2,3a{t}_2\right)$

Now,

$h=2a{t}_2^{2}k=3a{t}_2\frac{h}{2a}={\left(\frac{k}{3a}\right)}^{2}9a^{2}h=2a{k}^{2}9ax=2y^2$

Hence,

Option $\left(A\right)$ is correct answer.