Let (h, k) lie on the circle C: x^2 + y^2 = 4 | vedclass

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(C) Let the point $(h, k)$ on the circle $x^2 + y^2 = 4$ be represented as $(2 \cos 	heta, 2 \sin 	heta)$.Let the point $(x, y) = (2h + 1, 3k + 2)$.

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Let the point $(h, k)$ on the circle $x^2 + y^2 = 4$ be represented as $(2 \cos 	heta, 2 \sin 	heta)$.Let the point $(x, y) = (2h + 1, 3k + 2)$.Substituting $h = 2 \cos 	heta$ and $k = 2 \sin 	heta$, we get $x = 4 \cos 	heta + 1$ and $y = 6 \sin 	heta + 2$.Rearranging, we have $\cos 	heta = \frac{x - 1}{4}$ and $\sin 	heta = \frac{y - 2}{6}$.Using the identity $\cos^2 	heta + \sin^2 	heta = 1$, we get $(\frac{x - 1}{4})^2 + (\frac{y - 2}{6})^2 = 1$.This is the equation of an ellipse with semi-major axis $a = 6$ and semi-minor axis $b = 4$.The eccentricity $e$ is given by $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{16}{36} = 1 - \frac{4}{9} = \frac{5}{9}$.Therefore, $\frac{5}{e^2} = \frac{5}{5/9} = 9$.

Let $(h, k)$ lie on the circle $C: x^2 + y^2 = 4$ and the point $(2h + 1, 3k + 2)$ lie on an ellipse with eccentricity $e$. Then the value of $\frac{5}{e^2}$ is equal to . . . . . . .

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