If $x^2-y^2 = 1$, then which of the following is correct?

(A) $(x^2 - 1) (\frac{dy}{dx})^2=x^2$
(B) $(x^2-1)(\frac{d^2y}{dx^2})^2=x^2$
(C) $(x^2-1)^3(\frac{d^2y}{dx^2})^2=x^2$
(D) $(x^2-1)^3(\frac{d^2y}{dx^2})^2=1$

Choose the correct answer from the options given below.

(A) and (D) only

The correct answer is Option (3) → (A) and (D) only

Given relation

$x^2-y^2=1$

Differentiate w.r.t. $x$

$2x-2y\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{x}{y}$

Square both sides

$\left(\frac{dy}{dx}\right)^2=\frac{x^2}{y^2}$

Since $y^2=x^2-1$

$\left(\frac{dy}{dx}\right)^2=\frac{x^2}{x^2-1}$

Multiply both sides by $(x^2-1)$

$(x^2-1)\left(\frac{dy}{dx}\right)^2=x^2$

So option (A) is correct.

Now differentiate $\frac{dy}{dx}=\frac{x}{y}$ again

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

Substitute $\frac{dy}{dx}=\frac{x}{y}$

$\frac{d^2y}{dx^2}=\frac{y-\frac{x^2}{y}}{y^2}$

$=\frac{y^2-x^2}{y^3}$

Since $y^2=x^2-1$

$\frac{d^2y}{dx^2}=\frac{-1}{y^3}$

Square both sides

$\left(\frac{d^2y}{dx^2}\right)^2=\frac{1}{y^6}$

But $y^6=(x^2-1)^3$

So

$(x^2-1)^3\left(\frac{d^2y}{dx^2}\right)^2=1$

So option (D) is correct.

Correct options are (A) and (D)