Q84 / 150 · MHT CET 2026 (20 April Shift 1) · Mathematics

If $\sqrt{y + x} + \sqrt{y - x} = c$ and $\dfrac{dy}{dx} = K - \sqrt{\dfrac{y^2}{x^2} - 1}$ , then the value of $K$ is

  • A
    $\dfrac{-x}{y}$
  • B
    $\dfrac{-y}{x}$
  • C
    $\dfrac{x}{y}$
  • Correct answer
    D
    $\dfrac{y}{x}$

Solution

Given equation is $\sqrt{y + x} + \sqrt{y - x} = c$

Squaring both sides, we get:

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

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

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

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

Differentiating both sides with respect to $x$:

$\dfrac{dy}{dx} + \dfrac{1}{2\sqrt{y^2 - x^2}} \left( 2y \dfrac{dy}{dx} - 2x \right) = 0$

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

$\dfrac{dy}{dx} \left( 1 + \dfrac{y}{\sqrt{y^2 - x^2}} \right) = \dfrac{x}{\sqrt{y^2 - x^2}}$

$\dfrac{dy}{dx} \left( \dfrac{\sqrt{y^2 - x^2} + y}{\sqrt{y^2 - x^2}} \right) = \dfrac{x}{\sqrt{y^2 - x^2}}$

$\dfrac{dy}{dx} = \dfrac{x}{y + \sqrt{y^2 - x^2}}$

Rationalizing the denominator by multiplying the numerator and denominator by $y - \sqrt{y^2 - x^2}$:

$\dfrac{dy}{dx} = \dfrac{x(y - \sqrt{y^2 - x^2})}{(y + \sqrt{y^2 - x^2})(y - \sqrt{y^2 - x^2})}$

$\dfrac{dy}{dx} = \dfrac{x(y - \sqrt{y^2 - x^2})}{y^2 - (y^2 - x^2)}$

$\dfrac{dy}{dx} = \dfrac{x(y - \sqrt{y^2 - x^2})}{x^2}$

$\dfrac{dy}{dx} = \dfrac{y - \sqrt{y^2 - x^2}}{x}$

$\dfrac{dy}{dx} = \dfrac{y}{x} - \sqrt{\dfrac{y^2 - x^2}{x^2}}$

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

Comparing this with the given expression $\dfrac{dy}{dx} = K - \sqrt{\dfrac{y^2}{x^2} - 1}$, we get:

$K = \dfrac{y}{x}$

Answer: $\dfrac{y}{x}$

Hence, the correct answer is option: 'D'.

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