Test Index

Ellipse

Section: Mathematics

© examsnet.com

Question : 113 of 117

Marks: +1, -0

The ellipse

x^2+4 y^2=4

is inscribed in a rectangle aligned with the coordinate axex, which in turn is inscribed in another ellipse that passes through the point

(4,0)

. Then the equation of the ellipse is:

$x^2+12 y^2=16$
$4 x^2+48 y^2=48$
$4 x^2+64 y^2=48$
$x^2+16 y^2=16$

Solution:

The given ellipse is

\frac{x^2}{4}+\frac{y^2}{1}=1

So

A=(2,0)

and

B=(0,1)

If

P Q R S

is the rectangular in which it is inscribed, then

P=(2,1)

Let

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

be the ellipse circumscribing the rectangular

P Q R S

.

Then it passes through

P(2,1)

\therefore \;\frac{4}{a^2}+\;\frac{1}{b^2}=1 \;\; \dots (a)

Also, given that, it passes through

(4,0)

\;\therefore\;\frac{16}{a^2}+0=1 \Rightarrow a^2=16

\;\Rightarrow b^2=\frac{4}{3} [\;\text{substituting}\; a^2=16 \;\text{in}\; eq^n(a)]

\therefore

The required ellipse is

\;\frac{x^2}{16}+\;\frac{y^2}{\frac{4}{3}}=1

or

x^2+12 y^2=16

© examsnet.com

Go to Question:

Prev Question

Next Question