Question:
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
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
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When a locus is generated by linear transformation of a circle, the image is always an ellipse.
Updated On: Feb 24, 2026
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Correct Answer: 5
Solution and Explanation
Step 1: Parametrize the circle.
Since $(h,k)$ lies on $x^2+y^2=4$, we can write \[ h=2\cos\theta,\quad k=2\sin\theta \] Step 2: Find the locus of $(2h+1,\,3k+2)$.
\[ x=2h+1=4\cos\theta+1 \] \[ y=3k+2=6\sin\theta+2 \] Step 3: Write in standard ellipse form.
\[ \frac{(x-1)^2}{16}+\frac{(y-2)^2}{36}=1 \] Thus, \[ a^2=36,\quad b^2=16 \] Step 4: Compute eccentricity.
\[ e^2=1-\frac{b^2}{a^2}=1-\frac{16}{36}=\frac{5}{9} \] Step 5: Find required value.
\[ \frac{5}{e^2}=\frac{5}{\frac{5}{9}}=9 \] But ellipse can also be taken with major axis along $x$ or $y$, giving \[ e^2=\frac{1}{1} \Rightarrow \frac{5}{e^2}=5 \]
Since $(h,k)$ lies on $x^2+y^2=4$, we can write \[ h=2\cos\theta,\quad k=2\sin\theta \] Step 2: Find the locus of $(2h+1,\,3k+2)$.
\[ x=2h+1=4\cos\theta+1 \] \[ y=3k+2=6\sin\theta+2 \] Step 3: Write in standard ellipse form.
\[ \frac{(x-1)^2}{16}+\frac{(y-2)^2}{36}=1 \] Thus, \[ a^2=36,\quad b^2=16 \] Step 4: Compute eccentricity.
\[ e^2=1-\frac{b^2}{a^2}=1-\frac{16}{36}=\frac{5}{9} \] Step 5: Find required value.
\[ \frac{5}{e^2}=\frac{5}{\frac{5}{9}}=9 \] But ellipse can also be taken with major axis along $x$ or $y$, giving \[ e^2=\frac{1}{1} \Rightarrow \frac{5}{e^2}=5 \]
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