If CP and CD are a pair of semi-conjugate dia | vedclass

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(B) Let the coordinates of $P$ be $(a \cos 	heta, b \sin 	heta)$.Since $CD$ is the semi-conjugate diameter,the coordinates of $D$ are $(a \cos(	het

Vedclass · Accepted Answer

$a^{2}+b^{2}$

Vedclass · Answer

(B) Let the coordinates of $P$ be $(a \cos 	heta, b \sin 	heta)$.Since $CD$ is the semi-conjugate diameter,the coordinates of $D$ are $(a \cos(	heta + \frac{\pi}{2}), b \sin(	heta + \frac{\pi}{2})) = (-a \sin 	heta, b \cos 	heta)$.Now,$CP^{2} = a^{2} \cos^{2} 	heta + b^{2} \sin^{2} 	heta$.And $CD^{2} = (-a \sin 	heta)^{2} + (b \cos 	heta)^{2} = a^{2} \sin^{2} 	heta + b^{2} \cos^{2} 	heta$.Adding these,we get $CP^{2} + CD^{2} = a^{2}(\cos^{2} 	heta + \sin^{2} 	heta) + b^{2}(\sin^{2} 	heta + \cos^{2} 	heta)$.Since $\cos^{2} 	heta + \sin^{2} 	heta = 1$,we have $CP^{2} + CD^{2} = a^{2} + b^{2}$.

If $CP$ and $CD$ are a pair of semi-conjugate diameters of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,then $CP^{2}+CD^{2}=$

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