Question

If tangents are drawn from a point $P(2,0)$ to the curve $\sqrt{1+y^2}=\frac{x}{3}$ which meet the curve at $A$ and $B,$ then

• A. acute angle between the tangents is ${\tan }^{-1}\left(\frac{\sqrt{5}}{2}\right)$
• B. area of $△PAB$ is $\frac{5\sqrt{5}}{4}$ sq. units
• C. $AB=\sqrt{5}$
• D. $△PAB$ is equilateral

Answer 1

Answer: (A), (B), (C)

$AB=\sqrt{5}$

$\sqrt{1+y^2}=\frac{x}{3}$
Differentiating w.r.t. $x,$
$\frac{2y}{2\sqrt{1+y^2}}\left(\frac{dy}{dx}\right)=\frac{1}{3}$
$\Rightarrow \frac{dy}{dx}=\frac{\sqrt{1+y^2}}{3y}$
$\frac{dy}{dx}{|}_{(h,k)}=\frac{\sqrt{1+k^2}}{3k}$
$\therefore \frac{k-0}{h-2}=\frac{\sqrt{1+k^2}}{3k}$
$\Rightarrow \frac{3k^2}{h-2}=\sqrt{1+k^2}$
$\Rightarrow \frac{3(\frac{h^2}{9}-1)}{h-2}=\frac{h}{3}$
$\Rightarrow \frac{h^2-9}{h-2}=h$
$\Rightarrow h^2-9=h^2-2h$
$\Rightarrow h=\frac{9}{2}$
So, $k=\pm \frac{\sqrt{5}}{2}$

Let $AP$ make angle $\theta$ with $x-$axis in anticlockwise direction.
Then, acute angle between the tangents is $2\theta .$
Slope of $AP$ is $\tan \theta =\frac{1}{\sqrt{5}}$
$\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{\sqrt{5}}{2}$
$\Rightarrow 2\theta ={\tan }^{-1}\left(\frac{\sqrt{5}}{2}\right)$

$AB=\sqrt{5}$

Area of $△PAB$ is $\frac{1}{2}\times \sqrt{5}\times (\frac{9}{2}-2)=\frac{5\sqrt{5}}{4}$

$AP\ne AB$
Hence, $△PAB$ is not equilateral.