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: Represent a general point on the given circle

The circle is

x² + y² = 4

A general point on this circle can be written as:

P(h, k) = (2cosθ, 2sinθ)

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Step 2: Express the coordinates of point Q

Given,

Q = (2h + 1, 3k + 3)

Substituting h = 2cosθ and k = 2sinθ:

Q = (4cosθ + 1, 6sinθ + 3)

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Step 3: Eliminate the parameter θ to find the locus

From x-coordinate:

x − 1 = 4cosθ

⇒ cosθ = (x − 1)/4

From y-coordinate:

y − 3 = 6sinθ

⇒ sinθ = (y − 3)/6

Using the identity cos²θ + sin²θ = 1:

(x − 1)²/16 + (y − 3)²/36 = 1

This represents an ellipse.

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Step 4: Identify the semi-axes of the ellipse

Comparing with the standard form:

(x − h)²/b² + (y − k)²/a² = 1

we get:

a² = 36,   b² = 16

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Step 5: Compute the eccentricity

For an ellipse:

e² = 1 − b²/a²

e² = 1 − 16/36 = 5/9

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Step 6: Required value

5 / e² = 5 ÷ (5/9)

= 9

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Final Answer:

The required value is
9