Test Index

Ellipse

Section: Mathematics

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Question : 10 of 117

Marks: +1, -0

Let

(h, k)

lie on the circle

C : x^2+y^2=4

(2 h+1,3 k+2)

lie on an ellipse with eccentricity

e

. Then the value of

\; \frac{5}{e^2}

\_\_\_\_

[24 Jan 2026 Shift 2]

Your Answer:

Solution:

Concept:

A point on a circle is transformed to lie on an ellipse.

We derive the ellipse equation and compute its eccentricity

e

.

Explanation:

Since

(h,k)

lies on

x^2+y^2=4

, we have

h^2+k^2=4

.

Let

x=2h+1

and

y=3k+2

. Then

h=\frac{x-1}{2}

and

k=\frac{y-2}{3}

.

Substitute into

h^2+k^2=4

:

\left(\frac{x-1}{2}\right)^2+\left(\frac{y-2}{3}\right)^2=4

Simplify:

\frac{(x-1)^2}{4}+\frac{(y-2)^2}{9}=4

.

Divide by 4:

\frac{(x-1)^2}{16}+\frac{(y-2)^2}{36}=1

.

This ellipse has center

(1,2)

; semi-major axis

b=6

(since

36>16

) and semi-minor axis

a=4

.

Eccentricity

e=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{16}{36}}=\sqrt{\frac{20}{36}}=\sqrt{\frac{5}{9}}

.

Thus

e^2=\frac{5}{9}

.

Therefore

\frac{5}{e^2}=5\div\frac{5}{9}=9

.

Answer:

9

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