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MHTCET 20 Apr 2026 Shift 1 Paper

Question : 104 of 150

Marks: +1, -0

\sec ^{-1}\left(\frac{x^{2}+y^{2}}{x^{2}-y^{2}}\right)=2 a

y \frac{d y}{d x}=x \cdot f(a)

f\left(\frac{2 \pi}{3}\right)

Solution:

Concept:

Use the given inverse trigonometric equation to express

\frac{y^2}{x^2}

in terms of

a

, then differentiate to find

f(a)

Explanation:

Given:

\sec^{-1}\left(\frac{x^2+y^2}{x^2-y^2}\right)=2a

Taking secant on both sides:

\frac{x^2+y^2}{x^2-y^2}=\sec 2a

Divide numerator and denominator by

x^2

\frac{1+\frac{y^2}{x^2}}{1-\frac{y^2}{x^2}}=\sec 2a

Let

t=\frac{y^2}{x^2}

Then:

\frac{1+t}{1-t}=\sec 2a

Solving for

t

1+t=\sec 2a(1-t)

1+t=\sec 2a-t\sec 2a

t(1+\sec 2a)=\sec 2a-1

t=\frac{\sec 2a-1}{\sec 2a+1}

Using

\sec 2a=\frac{1}{\cos 2a}

t=\frac{1-\cos 2a}{1+\cos 2a}

Using the identity

\frac{1-\cos 2a}{1+\cos 2a}=\tan^2 a

\frac{y^2}{x^2}=\tan^2 a

So:

y^2=x^2\tan^2 a

Differentiate both sides with respect to

x

2y\frac{dy}{dx}=2x\tan^2 a

y\frac{dy}{dx}=x\tan^2 a

Comparing with

y\frac{dy}{dx}=x\cdot f(a)

f(a)=\tan^2 a

Now evaluate at

a=\frac{2\pi}{3}

f\left(\frac{2\pi}{3}\right)=\tan^2\left(\frac{2\pi}{3}\right)

Since

\tan\left(\frac{2\pi}{3}\right)=-\sqrt{3}

\tan^2\left(\frac{2\pi}{3}\right)=3

Thus:

f\left(\frac{2\pi}{3}\right)=3

Answer:

Option C:

3

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