x^2 a^2 -y^2 b^2 =1 is a solution of - Vidyadip

MCQ

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is a solution of

A
$\frac{d^2 y}{d x^2}+y x+\left(\frac{d y}{d x}\right)^2=0$
✓
$x y \frac{d^2 y}{d x^2}+2\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0$
C
$y \frac{d^2 y}{d x^2}+2\left(\frac{d y}{d x}\right)^2+y=0$
D
$x y \frac{d y}{d x}+y \frac{d^2 y}{d x^2}=0$

✓

Answer

Correct option: B.

$x y \frac{d^2 y}{d x^2}+2\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0$

$x y \frac{d^2 y}{d x^2}+2\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0$

Hint: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

... (1)

$\therefore \frac{1}{a^2} \times 2 x-\frac{1}{b^2} \times 2 y \frac{d y}{d x}=0$

$\therefore \frac{x}{a^2}-\frac{y}{b^2} \frac{d y}{d x}=0$

$\ldots$ (2)

and $\frac{1}{a^2} \times 1-\frac{1}{b^2}\left[y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2\right]=0$

Equations (1), (2) and (3) are consistent

$\therefore\left|\begin{array}{ccc}x^2 & -y^2 & 1 \\ x & -y \frac{d y}{d x} & 0 \\ 1 & -\left[y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2\right] & 0\end{array}\right|=0$

$\left.\therefore x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0\right]$

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