MHT CET · Maths · Ellipse
If \(\mathrm{CP}\) and \(\mathrm{CD}\) is a pair of semi-conjugate diameters of the ellipse \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1\), then
\(\mathrm{CP}^{2}+\mathrm{CD}^{2}=\)
- A \(\frac{a^{2}+b^{2}}{2}\)
- B \(a^{2}+b^{2}\)
- C \(a^{2}-b^{2}\)
- D \(\frac{a^{2}-b^{2}}{2}\)
Answer & Solution
Correct Answer
(B) \(a^{2}+b^{2}\)
Step-by-step Solution
Detailed explanation
The coordinates of \(\mathrm{P}\) and \(\mathrm{D}\) are \((a \cos \theta, b \sin \theta)\) and \(\left(a \cos \left(\frac{\pi}{2}+\theta\right), b \sin \left(\frac{\pi}{2}+\theta\right)\right)\) respectively.
\(\therefore C P^{2}+C D^{2}=a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta\)
\(\Rightarrow \quad C P^{2}+C D^{2}=a^{2}+b^{2}\)
\(\therefore C P^{2}+C D^{2}=a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta\)
\(\Rightarrow \quad C P^{2}+C D^{2}=a^{2}+b^{2}\)
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