Conic Sections Question 90
Question: The ellipse $ x^{2}+4y^{2}=4 $ is inscribed in a rectangle aligned with the coordinate axes, which is in turn inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is
Options:
A) $ x^{2}+16y^{2}=16 $
B) $ x^{2}+12y^{2}=16 $
C) $ 4x^{2}+48y^{2}=48 $
D) $ 4x^{2}+64y^{2}=48 $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ x^{2}+4y^{2}=4\Rightarrow \frac{x^{2}}{4}+\frac{y^{2}}{1}=1 $ So $ a=2, $
$ b=1, $ Thus P is (2, 1). The required ellipse is $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $
$ \Rightarrow \frac{x^{2}}{4^{2}}+\frac{y^{2}}{b^{2}}=1 $ The point (2, 1) lies on it. So $ \frac{4}{16}+\frac{1}{b^{2}}=1 $
$ \Rightarrow \frac{1}{b^{2}}=1-\frac{1}{4}=\frac{3}{4} $
$ \Rightarrow b^{2}=\frac{4}{3} $
$ \therefore \frac{x^{2}}{16}+\frac{y^{2}}{( \frac{4}{3} )}=1 $
$ \Rightarrow \frac{x^{2}}{16}+\frac{3y^{2}}{4}=1 $
$ \Rightarrow x^{2}+12y^{2}=16 $
BETA
