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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
- A \(2\)
- B \(\frac{4}{3}\)
- C \(\frac{1}{2}\)
- D \(\frac{3}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{3}\)
Step-by-step Solution
Detailed explanation
\(\left( b \right)\,\,\,\frac{{{x^2}}}{\alpha } + \frac{{{y^2}}}{4} = 1 \Rightarrow \frac{{2x}}{\alpha } + \frac{{2y}}{4}.\frac{{dy}}{{dx}} = 0\) \( \Rightarrow \frac{{dy}}{{dx}} = \frac{{ - 4x}}{{\alpha y}}\,\,\,\,\,\,...\left( i \right)\)…
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