Question

The ellipse $x^2+4y^2=4$ is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point $(4,0)$. Then the equation of the ellipse is

• A. $x^2+12y^2=16$
• B. $4x^2+48y^2=48$
• C. $4x^2+64y^2=48$
• D. $x^2+16y^2=16$

Answer 1

The correct option is A $x^2+12y^2=16$

Given Ellipse $\to x^2+4y^2=4$

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

So, our required ellipse passes through $(2,1),(2,-1),(-2,1),(-2,-1)$

we will check by putting these $4$ Pts. in each option.

$\left(B\right)$ $4x^2+48y^2=48$

For $(2,1)$

$4(2{)}^2+48(1{)}^2=48$

$16+48\ne 48$

(wrong option)

$\left(C\right)$ $4x^2+64y^2=48$

For $(2,1)$

$4(2{)}^2+64(1{)}^2=48$

$16+64\ne 48$

(wrong option)

$D)$ $x^2+16y^2=16$

For $(2,1),\left(2^2\right)+16(1{)}^2=16$

$4+16\ne 16$

(wrong option)

$\left(A\right)$ $x^2+12y^2=16$

For $(2,),(2{)}^2+12(1{)}^2=4+12=16$

$(-2,1),2^2+12(-1{)}^2=16$

$(-2,1),(-2{)}^2+12(1{)}^2=16$

$(-2,-1),(-2{)}^2+12(-1{)}^2=16$

(correct option)

Ans. is $\left(A\right)x^2+12y^2=16$