Question

If $x=\sin t$, $y=\sin kt$ then value of $(1-x^2)y_2-x{y}_1$ is

• A. $-k^{2}y$
• B. $k^{2}y$
• C. $k{y}^2$
• D. $-k{y}^2$

Answer 1

The correct option is A $-k^{2}y$

$\because$ $x=\sin t$, $y=\sin kt$

$\Rightarrow \frac{dx}{dt}=\cos t,\frac{dy}{dt}=k\cos kt$
$\therefore$ $y_1=\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{k\cos kt}{\cos t}$
$\Rightarrow$ $y_1\cos t=k\cos kt$

On squaring both sides, we have

$y_1^2{\cos }^{2}t=k^2{\cos }^{2}kt$
$y_1^2(1-{\sin }^{2}t)=k^2(1-{\sin }^{2}kt)$
$\Rightarrow$ $y_1^2(1-x^2)=k^2(1-y^2)$
Differentiating both sides w.r to x, we have
$(1-x^2)\left(2y_1{y}_2\right)+y{{}_1}^2(-2x)=k^2(-2y{y}_1)$
$\Rightarrow$ $(1-x^2)y_2-x{y}_1=-k^{2}y$