Question

If $\sqrt{x+y}+\sqrt{y-x}=c,$ then $\frac{d^{2}x}{d{y}^2}$ is equal to

• A. $\frac{2}{c}$
• B. $-\frac{2}{c^2}$
• C. $\frac{2}{c^2}$
• D. none of these

Answer 1

Answer: (C)

$\frac{2}{c^2}$

$\sqrt{x+y}+\sqrt{y-x}=c$ ...(1)

Rationalising we get

$\frac{(y+x)-(y-x)}{\sqrt{x+y}-\sqrt{y-x}}=c$

We know that $(y+x)-(y-x)=2x$

we get

$\sqrt{x+y}-\sqrt{y-x}=\frac{2x}{c}$ ...(2)

From (1) and (2)

$2\sqrt{y+x}=c+\frac{2x}{c}$

$\Rightarrow 4(y+x)=c^2+\frac{4x^2}{c^2}+4x$

$\Rightarrow 4y=c^2+\frac{4x^2}{c^2}\Rightarrow 4\frac{dy}{dx}=4.\frac{2x}{c^2}$

$\Rightarrow \frac{dy}{dx}=\frac{2x}{c^2}\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{2}{c^2}$