Question

Two tangents to parabola $y^2=4ax$ have inclinations ${\theta }_1$ and ${\theta }_2$ with $x$-axis such that ${\tan }^2{\theta }_1+{\tan }^2{\theta }_2=k$ then the locus of the point of intersection is

• A. $y=kx$
• B. $y^2=k{x}^2+2ax$
• C. $y^2=2ax$
• D. $y^2=2a$

Answer 1

The correct option is B $y^2=k{x}^2+2ax$

Equations of tangents are:

$y{t}_1=x+a{t}_1^2$

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

we have,

$y=a(t_1+t_2),x=a{t}_1{t}_2$

$\tan {\theta }_1=\frac{1}{t_1}$

$\tan {\theta }_2=\frac{1}{t_2}$

$h=a{t}_1{t}_2$

$h^2=a^2{t}_1^2{t}_2^2$

$h^2=a^2\frac{1}{\tan {\theta }_1^2}\frac{1}{\tan {\theta }_2^2}$

$\frac{a^2}{h^2}=\tan {\theta }_1^2\tan {\theta }_2^2$

$k=a(t_1+t_2)$

$k^2=a^2(t_1^2+t_2^2+2t_1{t}_2)$

$=a^2(\frac{1}{\tan {\theta }_1^2}+\frac{1}{\tan {\theta }_2^2}+2\frac{1}{\tan {\theta }_1^2}\frac{1}{\tan {\theta }_2^2})$

$k^2=a^2(\frac{k{h}^2}{a^2}+\frac{2h}{a})$

$k^2=k{h}^2+2ah$

$y^2=k{x}^2+2ax$