Q. 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 \(\dfrac{5}{e^2}\) is equal to
Explanation
Given \((h,k)\) lies on the circle
\[
h^2+k^2=4
\]
The transformed point is \((2h+1,3k+2)\). Subtracting the translation,
let
\[
X=2h,\quad Y=3k
\]
Then
\[
\frac{X^2}{4}+\frac{Y^2}{9}=h^2+k^2=4
\]
Dividing throughout by \(4\),
\[
\frac{X^2}{16}+\frac{Y^2}{36}=1
\]
This represents an ellipse with
\[
a^2=36,\quad b^2=16
\]
So,
\[
e^2=1-\frac{b^2}{a^2}=1-\frac{16}{36}=\frac{5}{9}
\]
Hence,
\[
\frac{5}{e^2}=\frac{5}{5/9}=9
\]
Therefore, the correct answer is 9.