JEE Main 12 Jan 2019 Evening Question 1
Question: Let s and $ S’ $ be the foci of an ellipse and B be any one of the extremities of its minor axis. If $ \Delta S’BS $ is a right angled triangle with right angle at B and area $ (\Delta S’BS)=8sq. $ units, then the length of altos rectum of the ellipse is [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) 2
B) 4
C) $ 4\sqrt{2} $
D) $ 2\sqrt{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Since $ \angle S’BS=90^{o} $
$ \therefore $ Slope of $ S’B\times $ Slope of $ SB=-1 $
$ \Rightarrow $ $ \frac{-b}{-ae}\times \frac{-b}{ae}=-1\Rightarrow b^{2}=a^{2}e^{2} $ ?(i) Now, area of $ \Delta S’BS=8 $
$ \Rightarrow $ $ \frac{1}{2}\times 2ae\times b=8 $
$ \Rightarrow $ $ ae\times b=8 $
$ \Rightarrow $ $ a^{2}e^{2}\times b^{2}=64 $
$ \Rightarrow $ $ {{(a^{2}e^{2})}^{2}}=64 $ [Using (i)]
$ \Rightarrow $ $ a^{2}e^{2}=8\Rightarrow ae=\pm 2\sqrt{2} $
$ \therefore $ $ b=2\sqrt{2} $ Again, $ a^{2}=a^{2}e^{2}+b^{2} $ $ [ \because e=\sqrt{1-\frac{b^{2}}{a^{2}}} ] $
$ \Rightarrow $ $ a^{2}=8+8=16\Rightarrow a=\pm 4 $
$ \therefore $ Length of latus rectum $ =\frac{2b^{2}}{a}=\frac{2\times 8}{4}=4 $
