Question

The sum of the intercepts of the tangent to the curve $\sqrt{x}+\sqrt{y}=3$ on the co-ordinate axes where it meets the curve $y=e^x$

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

The correct option is A 9

Let both curves meets at $(x_1,y_1)$

$\because \sqrt{x}+\sqrt{y}=3$

so , $\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=\frac{\sqrt{y_1}}{\sqrt{x_1}}$

so, equation of tangent $y-y_1=\frac{\sqrt{y_1}}{\sqrt{x_1}}(x-x_1)$

$xintercept=x_1+\sqrt{x_1{y}_1}$

$yintercept=y_1+\sqrt{x_1{y}_1}$

Let the point P be $(x_1,y_1)$

so , $\sqrt{x_1}+\sqrt{y_1}=3$

sum of intercepts $(\sqrt{x_1}+\sqrt{y_1}{)}^2=9$