Q59 / 150 · MHT CET 2020 (15 Oct Shift 2) · Mathematics

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
    $\frac{a^{2}+b^{2}}{2}$
  • Correct answer
    B
    $a^{2}+b^{2}$
  • C
    $a^{2}-b^{2}$
  • D
    $\frac{a^{2}-b^{2}}{2}$

Solution

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. $\therefore C P^{2}+C D^{2}=a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta$ $\Rightarrow \quad C P^{2}+C D^{2}=a^{2}+b^{2}$

Hence, the correct answer is option: 'B'.

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