question_answer

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]

A) 2         

B)                           4    

C) \[4\sqrt{2}\]                              

D)   \[2\sqrt{2}\]

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,\[{{a}^{2}}={{a}^{2}}{{e}^{2}}+{{b}^{2}}\]         \[\left[ \because e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}} \right]\] \[\Rightarrow \]\[{{a}^{2}}=8+8=16\Rightarrow a=\pm 4\] \[\therefore \]Length of latus rectum\[=\frac{2{{b}^{2}}}{a}=\frac{2\times 8}{4}=4\]