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Ellipse

Section: Mathematics

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Question : 4 of 16

Marks: +1, -0

Let

A_1

be the area of the given ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

A_2

be the area of the region bounded by the curve which is the locus of mid point of the line segment joining the focus of the ellipse and a point

P

on the given ellipse, then

A_1: A_2

=

[AP EAPCET 22 May 2025 S2]

$3: 2$
a:b
4:1
2a:3b

Solution:

Area of ellipse

A_1=\pi a b

Let

(x_1, y_1)

be a point on the ellipse Let

(a e, 0)

be the focus Let

(h, k)

be the mid-point.

h=\frac{x_1+a e}{2}, k=\frac{y_1}{2}

x_1=2 h-a e, y_1=2 k

\Rightarrow \frac{(2 h-a e)^2}{a^2}+\frac{(2 k)^2}{b^2}=1

\Rightarrow \frac{\left(h-\frac{a e}{2}\right)^2}{\left(\frac{a}{2}\right)^2}+\frac{k^2}{\left(\frac{b}{2}\right)^2}=1

\Rightarrow \text{ Semi-major axis } =\frac{a}{2}

Semi-minor axis

=\frac{b}{2}

The area of the locus ellipse

A_2=\pi \cdot \frac{a}{2} \cdot \frac{b}{2}=\frac{a b \pi}{4}

\Rightarrow \frac{A_1}{A_2}=\frac{\pi a b}{\frac{\pi a b}{4}}=\frac{4}{1}

\Rightarrow A_1: A_2=4: 1

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