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JEE Mains · Maths · STD 12 - 6. Application of derivatives

If the curves \(\frac{{{x^2}}}{\alpha } + \frac{{{y^2}}}{4} = 1\) and \({y^3} = 16x\) intersect at right angles, then a value of \(\alpha \) is

  1. A \(2\)
  2. B \(\frac{4}{3}\)
  3. C \(\frac{1}{2}\)
  4. D \(\frac{3}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{4}{3}\)

Step-by-step Solution

Detailed explanation

\(\left( b \right)\,\,\,\frac{{{x^2}}}{\alpha } + \frac{{{y^2}}}{4} = 1 \Rightarrow \frac{{2x}}{\alpha } + \frac{{2y}}{4}.\frac{{dy}}{{dx}} = 0\) \( \Rightarrow \frac{{dy}}{{dx}} = \frac{{ - 4x}}{{\alpha y}}\,\,\,\,\,\,...\left( i \right)\)…