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MHTCET 20 Apr 2026 Shift 1 Paper

Question : 121 of 150

Marks: +1, -0

\sqrt{y+x}+\sqrt{y-x}=c

\frac{d y}{d x}=K-\sqrt{\frac{y^{2}}{x^{2}}-1}

K

Solution:

Concept:

Differentiate the given equation implicitly and compare it with the standard form to find the value of

K

Explanation:

Given:

\sqrt{y+x}+\sqrt{y-x}=c

Squaring both sides gives

2y+2\sqrt{y^{2}-x^{2}}=c^{2}

So,

y+\sqrt{y^{2}-x^{2}}=\frac{c^{2}}{2}

, which is a constant.

Differentiating with respect to

x

\frac{dy}{dx}+\frac{y\frac{dy}{dx}-x}{\sqrt{y^{2}-x^{2}}}=0

Let

p=\frac{dy}{dx}

. Then

p+\frac{yp-x}{\sqrt{y^{2}-x^{2}}}=0

This simplifies to

p\left(y+\sqrt{y^{2}-x^{2}}\right)=x

Hence

p=\frac{x}{y+\sqrt{y^{2}-x^{2}}}

Rationalising the denominator:

p=\frac{x\left(y-\sqrt{y^{2}-x^{2}}\right)}{x^{2}}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}

Thus

\frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}

Comparing with

\frac{dy}{dx}=K-\sqrt{\frac{y^{2}}{x^{2}}-1}

, we get

K=\frac{y}{x}

Answer:

K=\frac{y}{x}

, so the correct option is D.

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