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

• A. 9
• B. 5
• C. 3
• D. 6

Hint:

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

Answer 1

The Correct Option is A

The problem involves determining the eccentricity of an ellipse derived from the coordinates of a point that lies on the ellipse. Let's solve this step-by-step:

1. P(h,k) is a point on the circle x^2 + y^2 = 4. The equation of this circle can also be written as (h^2 + k^2 = 4).
2. Consider the point Q(2h+1, 3k+3). Substitute (h,k) from the circle's equation into the coordinates of Q.
3. The point Q(2h+1, 3k+3) lies on an ellipse, so we need the general form of the ellipse to find its parameters. Substituting and simplifying using the circle equation h^2 + k^2 = 4 will help us.
4. Substitution leads us to: \[ h = \frac{x - 1}{2}, \quad k = \frac{y - 3}{3} \] Plugging these into the circle's equation: \[ \left(\frac{x - 1}{2}\right)^2 + \left(\frac{y - 3}{3}\right)^2 = 4 \] \rightarrow (x - 1)^2/4 + (y - 3)^2/9 = 4 \]
5. Rearrange to the standard ellipse equation: \[ \frac{(x-1)^2}{16} + \frac{(y-3)^2}{36} = 1 \] This represents an ellipse centered at (1,3), with semi-major axis 6 and semi-minor axis 4.
6. The eccentricity (e) of an ellipse is given by: \[ e = \sqrt{1 - \left(\frac{b^2}{a^2}\right)} \] where a = 6 and b = 4.
7. Compute the eccentricity: \[ e = \sqrt{1 - \frac{4^2}{6^2}} = \sqrt{1 - \frac{16}{36}} = \sqrt{\frac{20}{36}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \] \]
8. We need to find \frac{5}{e^2}: \[ e^2 = \frac{5}{9} \Rightarrow \frac{5}{e^2} = \frac{5}{{5/9}} = 9 \]
9. Therefore, the value of \dfrac{5}{e^2} is 9.

Thus, the correct answer is 9.