Q63 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
If $x=2 a \sin ^{-1}\left(\sqrt{\dfrac{y}{2 a}}\right)-\sqrt{2 a y-y^2}$ then $\dfrac{d y}{d x}=$
- A$\sqrt{\dfrac{y}{2 a-y}}$
- B$\dfrac{\sqrt{2 a-y}}{y}$
- C$\dfrac{1}{2}\left(\sqrt{\dfrac{2 a-y}{y}}\right)$
- Correct answerD$\sqrt{\dfrac{2 a-y}{y}}$
Solution
Given $x = 2a \sin^{-1}\left(\sqrt{\dfrac{y}{2a}}\right) - \sqrt{2ay - y^2}$.
Let $y = 2a \sin^2 \theta$. Then $\sqrt{\dfrac{y}{2a}} = \sin \theta$, so $\theta = \sin^{-1}\left(\sqrt{\dfrac{y}{2a}}\right)$.
Also, $\sqrt{2ay - y^2} = \sqrt{2a(2a \sin^2 \theta) - (2a \sin^2 \theta)^2} = \sqrt{4a^2 \sin^2 \theta (1 - \sin^2 \theta)} = \sqrt{4a^2 \sin^2 \theta \cos^2 \theta} = 2a \sin \theta \cos \theta = a \sin(2\theta)$.
Substituting these into the expression for $x$:
$x = 2a \theta - a \sin(2\theta)$.
Differentiating $x$ with respect to $\theta$:
$\dfrac{dx}{d\theta} = 2a - 2a \cos(2\theta) = 2a(1 - \cos(2\theta)) = 2a(2 \sin^2 \theta) = 4a \sin^2 \theta$.
Differentiating $y = 2a \sin^2 \theta$ with respect to $\theta$:
$\dfrac{dy}{d\theta} = 2a(2 \sin \theta \cos \theta) = 4a \sin \theta \cos \theta$.
Using the chain rule $\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta}$:
$\dfrac{dy}{dx} = \dfrac{4a \sin \theta \cos \theta}{4a \sin^2 \theta} = \dfrac{\cos \theta}{\sin \theta} = \cot \theta$.
Since $\sin^2 \theta = \dfrac{y}{2a}$, we have $\cos^2 \theta = 1 - \dfrac{y}{2a} = \dfrac{2a - y}{2a}$.
Therefore, $\cot^2 \theta = \dfrac{\cos^2 \theta}{\sin^2 \theta} = \dfrac{(2a - y)/(2a)}{y/(2a)} = \dfrac{2a - y}{y}$.
Thus, $\dfrac{dy}{dx} = \sqrt{\dfrac{2a - y}{y}}$.
Answer: $\sqrt{\dfrac{2 a-y}{y}}$
Let $y = 2a \sin^2 \theta$. Then $\sqrt{\dfrac{y}{2a}} = \sin \theta$, so $\theta = \sin^{-1}\left(\sqrt{\dfrac{y}{2a}}\right)$.
Also, $\sqrt{2ay - y^2} = \sqrt{2a(2a \sin^2 \theta) - (2a \sin^2 \theta)^2} = \sqrt{4a^2 \sin^2 \theta (1 - \sin^2 \theta)} = \sqrt{4a^2 \sin^2 \theta \cos^2 \theta} = 2a \sin \theta \cos \theta = a \sin(2\theta)$.
Substituting these into the expression for $x$:
$x = 2a \theta - a \sin(2\theta)$.
Differentiating $x$ with respect to $\theta$:
$\dfrac{dx}{d\theta} = 2a - 2a \cos(2\theta) = 2a(1 - \cos(2\theta)) = 2a(2 \sin^2 \theta) = 4a \sin^2 \theta$.
Differentiating $y = 2a \sin^2 \theta$ with respect to $\theta$:
$\dfrac{dy}{d\theta} = 2a(2 \sin \theta \cos \theta) = 4a \sin \theta \cos \theta$.
Using the chain rule $\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta}$:
$\dfrac{dy}{dx} = \dfrac{4a \sin \theta \cos \theta}{4a \sin^2 \theta} = \dfrac{\cos \theta}{\sin \theta} = \cot \theta$.
Since $\sin^2 \theta = \dfrac{y}{2a}$, we have $\cos^2 \theta = 1 - \dfrac{y}{2a} = \dfrac{2a - y}{2a}$.
Therefore, $\cot^2 \theta = \dfrac{\cos^2 \theta}{\sin^2 \theta} = \dfrac{(2a - y)/(2a)}{y/(2a)} = \dfrac{2a - y}{y}$.
Thus, $\dfrac{dy}{dx} = \sqrt{\dfrac{2a - y}{y}}$.
Answer: $\sqrt{\dfrac{2 a-y}{y}}$
Hence, the correct answer is option: 'D'.