Question

The point of contact of the tangent to the parabola $y^2=9x$ which passes through the point $(4,10)$ and makes an angle $\theta$ with the axis of the parabola such that $\tan \theta >2$ is

• A. $(\frac{4}{9},2)$
• B. $(36,18)$
• C. $(4,6)$
• D. $(\frac{1}{4},\frac{3}{2})$

Answer 1

The correct option is A $(\frac{4}{9},2)$

Given parabola is $y^2=9x$

$\Rightarrow a=\frac{9}{4}$

Equation of a tangent to $y^2=9x$ at $(\frac{{9t}^2}{4},\frac{9t}{2})$ is

$ty=x+\frac{{9t}^2}{4}$, which passes through $(4,10)$
$\Rightarrow 40t=16+{9t}^2\Rightarrow {9t}^2-40t+16=0\Rightarrow t=\frac{4}{9}$ or $4$
But the slope $\frac{1}{t}$ of the tangent is $>2$
$\Rightarrow t<\frac{1}{2}$ or $t=\frac{4}{9}$
and the point of contact is $(\frac{4}{9},2)$