Question

If $P(h,k)$ is a variable point on $x^2 + y^2 = 4$ and $Q(2h+1,\,3k+3)$ always lies on an ellipse, if eccentricity of the ellipse is $e$, then $\dfrac{5}{e^2}$ is equal to

Hint:

Always parametrize the given locus first and then transform coordinates to find the new locus.

Answer 1

Correct Answer: 9

Step 1: Parametrize the given circle.
Since $P(h,k)$ lies on $x^2 + y^2 = 4$, let \[ P = (2\cos\theta,\,2\sin\theta) \] Step 2: Find coordinates of point $Q$.
\[ Q = (2h+1,\,3k+3) \] Substituting values of $h$ and $k$, \[ Q = (4\cos\theta + 1,\,6\sin\theta + 3) \] Step 3: Write the locus of $Q$.
\[ \frac{(x-1)^2}{16} + \frac{(y-3)^2}{36} = 1 \] This is the equation of an ellipse.
Step 4: Identify $a^2$ and $b^2$.
\[ a^2 = 36,\quad b^2 = 16 \] Step 5: Find eccentricity.
\[ e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{16}{36} = \frac{5}{9} \] Step 6: Final calculation.
\[ \frac{5}{e^2} = \frac{5}{\frac{5}{9}} = 9 \] Final conclusion.
The required value is 9.