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 \(\alpha^2+\beta^2=\)
- A 10
- B 9
- C 28
- D 19
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…
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