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

वक्रों \(\frac{ x ^{2}}{ a ^{2}}+\frac{ y ^{2}}{ b ^{2}}=1\) तथा \(x ^{2}+ y ^{2}= ab , a > b\), का एक प्रतिच्छेदन कोण है

  1. A \(\tan ^{-1}\left(\frac{a+b}{\sqrt{a b}}\right)\)
  2. B \(\tan ^{-1}\left(\frac{a-b}{2 \sqrt{a b}}\right)\)
  3. C \(\tan ^{-1}\left(\frac{\mathrm{a}-\mathrm{b}}{\sqrt{\mathrm{ab}}}\right)\)
  4. D \(\tan ^{-1}(2 \sqrt{a b})\)
Verified Solution

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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