TS EAMCET · Maths · Application of Derivatives
If the curves \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and \(\frac{x^2}{25}+\frac{y^2}{16}=1\) cut each other orthogonally, then \(a^2-b^2\) equals to
- A \(9\)
- B \(400\)
- C \(75\)
- D \(41\)
Answer & Solution
Correct Answer
(A) \(9\)
Step-by-step Solution
Detailed explanation
We know, two curves \(\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}=1\) and \(\frac{x^2}{a_2^2}+\frac{y^2}{b_2^2}=1\) cut orthogonally. Then, \(a_1^2-a_2^2=b_1^2-b_2^2\) Here, equation of curves are \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text { and } \frac{x^2}{25}+\frac{y^2}{16}=1 \]…
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