Search any question & find its solution
Question:
Answered & Verified by Expert
Normals are drawn from the point $\mathrm{P}(8,0)$ to the parabola $y^2=12 x$. If $\theta$ is the acute angle between two nonhorizontal normals among them, then $\tan \theta=$
Options:
Solution:
2452 Upvotes
Verified Answer
The correct answer is:
$2 \sqrt{6}$
$y^2=12 x, a=3$
Equation of normal is
$$
y=m x-2 a m-a m^3 \Rightarrow y=m x-6 m-3 m^3
$$
$\because(8,0)$ lies on normal
$$
\begin{aligned}
& 8 m-6 m-3 m^3=0 \Rightarrow m\left(2-3 m^2\right)=0 \\
& m=0, \pm \sqrt{\frac{2}{3}} \\
\therefore \quad & \tan \theta=\left(\frac{\sqrt{\frac{2}{3}}+\sqrt{\frac{2}{3}}}{1-\frac{2}{3}}\right)=\frac{2 \sqrt{2}}{\sqrt{3}} \times 3=2 \sqrt{6} \\
\therefore \quad & \tan \theta=2 \sqrt{6} .
\end{aligned}
$$
Equation of normal is
$$
y=m x-2 a m-a m^3 \Rightarrow y=m x-6 m-3 m^3
$$
$\because(8,0)$ lies on normal
$$
\begin{aligned}
& 8 m-6 m-3 m^3=0 \Rightarrow m\left(2-3 m^2\right)=0 \\
& m=0, \pm \sqrt{\frac{2}{3}} \\
\therefore \quad & \tan \theta=\left(\frac{\sqrt{\frac{2}{3}}+\sqrt{\frac{2}{3}}}{1-\frac{2}{3}}\right)=\frac{2 \sqrt{2}}{\sqrt{3}} \times 3=2 \sqrt{6} \\
\therefore \quad & \tan \theta=2 \sqrt{6} .
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.