JEE Mains · Maths · STD 12 - 6. Application of derivatives
An angle of intersection of the curves, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) and \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{ab}, \mathrm{a}>\mathrm{b}\), is :
- A \(\tan ^{-1}\left(\frac{a+b}{\sqrt{a b}}\right)\)
- B \(\tan ^{-1}\left(\frac{a-b}{2 \sqrt{a b}}\right)\)
- C \(\tan ^{-1}\left(\frac{\mathrm{a}-\mathrm{b}}{\sqrt{\mathrm{ab}}}\right)\)
- D \(\tan ^{-1}(2 \sqrt{a b})\)
Answer & Solution
Correct Answer
(C) \(\tan ^{-1}\left(\frac{\mathrm{a}-\mathrm{b}}{\sqrt{\mathrm{ab}}}\right)\)
Step-by-step Solution
Detailed explanation
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, x^{2}+y^{2}=a b\) \(\frac{2 x_{1}}{a^{2}}+\frac{2 y_{1} y^{\prime}}{b^{2}}=0\) \(\Rightarrow y_{1}^{\prime}=\frac{-x_{1}}{a^{2}} \frac{b^{2}}{y_{1}}...(1)\) \(\therefore 2 x_{1}+2 y_{1} y^{\prime}=0\)…
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