Question

The sum of intercepts of any tangent on the curve $\sqrt{x}+\sqrt{y}=2$ is

• A. $2\sqrt{2}$
• B. $2$
• C. $8$
• D. $4$

Answer 1

The correct option is D $4$

$\sqrt{x}+\sqrt{y}=2$
Let $x=4{\cos }^4\theta ,y=4{\sin }^4\theta$
$\frac{dx}{d\theta }=-16{\cos }^3\theta \sin \theta \frac{dy}{d\theta }=16{\sin }^3\theta \cos \theta \Rightarrow \frac{dy}{dx}=-{\tan }^2\theta$
So, the equation of tangent is,
$y-4{\sin }^4\theta =-{\tan }^2\theta (x-4{\cos }^4\theta )$

For $x-$ intercept, put $y=0$
$x=\frac{4{\sin }^4\theta }{{\tan }^2\theta }+4{\cos }^4\theta =4{\sin }^2\theta {\cos }^2\theta +4{\cos }^4\theta =4{\cos }^2\theta$

For $y-$ intercept, put $x=0$
$y=4{\sin }^4\theta +4{\cos }^4\theta \cdot {\tan }^2\theta =4{\sin }^4\theta +4{\sin }^2\theta {\cos }^2\theta =4{\sin }^2\theta$

Therefore, the sum of intercept
$=4{\cos }^2\theta +4{\sin }^2\theta =4$