Assertion (A): If the tangent and normal to t | vedclass

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(A) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$,so $a^2 = 16$ and $b^2 = 9$.The tangent at $P(a \cos 	heta, b \sin 	heta)$ i

Vedclass · Accepted Answer

Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.

Vedclass · Answer

(A) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$,so $a^2 = 16$ and $b^2 = 9$.The tangent at $P(a \cos 	heta, b \sin 	heta)$ is $\frac{x \cos 	heta}{a} + \frac{y \sin 	heta}{b} = 1$. It meets the major axis $(y=0)$ at $Q(\frac{a}{\cos 	heta}, 0)$.The normal at $P$ is $\frac{ax}{\cos 	heta} - \frac{by}{\sin 	heta} = a^2 - b^2$. It meets the major axis at $R(\frac{(a^2 - b^2) \cos 	heta}{a}, 0)$.The distance $QR = \left| \frac{a}{\cos 	heta} - \frac{(a^2 - b^2) \cos 	heta}{a} ight| = \left| \frac{a^2 - (a^2 - b^2) \cos^2 	heta}{a \cos 	heta} ight| = \left| \frac{a^2 \sin^2 	heta + b^2 \cos^2 	heta}{a \cos 	heta} ight|$.Thus,Reason $(R)$ is true.For $a=4, b=3, 	heta = \frac{\pi}{3}$,$QR = \left| \frac{16 \cdot \frac{3}{4} + 9 \cdot \frac{1}{4}}{4 \cdot \frac{1}{2}} ight| = \left| \frac{12 + 2.25}{2} ight| = \frac{14.25}{2} = \frac{57}{8}$.Thus,Assertion $(A)$ is also true and $(R)$ is the correct explanation.

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