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AP EAMCET · Maths · Application of Derivatives

If the tangent drawn at the point \((\alpha, \beta)\) on the curve \(x^{2 / 3}+y^{2 / 3}=4\) is parallel to the line \(\sqrt{3} x+y=1\), then \(\alpha^2+\beta^2=\)

  1. A 10
  2. B 9
  3. C 28
  4. D 19
Verified Solution

Answer & Solution

Correct Answer

(C) 28

Step-by-step Solution

Detailed explanation

Differentiate the curve: \(x^{2 / 3}+y^{2 / 3}=4\) \(\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\left(\frac{y}{x}\right)^{1/3}\) Slope of tangent at \((\alpha, \beta)\) is \(m_t = -\left(\frac{\beta}{\alpha}\right)^{1/3}\) Slope of line…