Question
Question: The ellipse \[{{\text{E}}_1}:\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{4} = 1\] is inscribed in a rectang...
The ellipse E1:9x2+4y2=1 is inscribed in a rectangle R whose sides are parallel to the coordinate axis. Another ellipse E2 is passing through the point (0, 4) circumscribing the rectangle R. The eccentricity of the rectangle E2 is
A. 22
B. 23
C. 21
D. 43
Solution
Here, first find the vertices of the ellipse E1 by comparing it with the standard equation. So you will get vertices of the rectangle. Assume the equation of ellipse E2 and this ellipse must satisfy the point of the rectangle. Using points of the rectangle in second, find the values of its vertices and then apply formula to find eccentricity.
Complete step-by-step answer:
Given equation of ellipse E1:9x2+4y2=1
By comparing this equation as standard equation of ellipse i.e., a2x2+b2y2=1 in which ellipse touches (± a,0) and y axis at (0,± b).
Therefore, the ellipse E1 touches x-axis at (± 3,0) and y axis at (0,± 2).
Also given that the ellipse is inscribed in rectangle R whose sides are parallel to the coordinate axis, then the vertices of the rectangle are (±3,±2).
Let the equation of ellipse E2:a2x2+b2y2=1
The ellipse circumscribes the rectangle R, so the vertices of rectangle lie on ellipse E2
Therefore we get a29+b24=1
Given that, the ellipse E2 passes through the point (0, 4).
So we get b = 4 and a2=12
For ellipse, we know thata2=b2(1−e2), where e is the eccentricity of the ellipse.
⇒ 12=16(1−e2) ⇒1−e2=43⇒e2=41⇒e=21
Therefore, the eccentricity of the rectangle E2 is 21.
Note: In these types of questions, first draw the clear diagram to see the given statement geometrically. Also all intersection points should be clearly visible. Always compare the given equation of ellipse with the standard equation and also assume any equation of ellipse as in the form of the standard equation. Also if any point is on the ellipse then it must satisfy the equation of ellipse use this fact to solve the problem.