Q66 / 155 · WBJEE 2022 · Mathematics
The point of contact of the tangent to the parabola $y^2=9 x$ which passes through the point $(4,10)$ and makes an angle $\theta$ with the positive side of the axis of the parabola where $\tan \theta>2$, is
- Correct answerA$\left(\frac{4}{9}, 2\right)$
- B$(4,6)$
- C$(4,5)$
- D$\left(\frac{1}{4}, \frac{1}{6}\right)$
Solution
$$
\tan \theta=\frac{1}{t}>2
$$
Equation of tangent at $\mathrm{t}$
$$
\begin{aligned}
&\left.\mathrm{yt}=\mathrm{x}+\frac{9}{4} \mathrm{t}^2 \text { (passes through }(4,10)\right) \Rightarrow 10 \mathrm{t}=4+\frac{9}{4} \mathrm{t}^2 \Rightarrow 9 \mathrm{t}^2-40 \mathrm{t}+16=0 \Rightarrow \mathrm{t}=4, \frac{4}{9}\left(\frac{1}{\mathrm{t}}>2\right) \Rightarrow \mathrm{t}=\frac{4}{9} \\
&\therefore \text { point }\left(\mathrm{at}^2, 2 \mathrm{at}\right) \Rightarrow\left(\frac{4}{9}, 2\right)
\end{aligned}
$$
Hence, the correct answer is option: 'A'.