Conic Sections Question 103
Question: An ellipse has eccentricity $ \frac{1}{2} $ and one focus at the point $ P( \frac{1}{2},\ 1 ) $ . Its one directrix is the common tangent nearer to the point P, to the circle $ x^{2}+y^{2}=1 $ and the hyperbola $ x^{2}-y^{2}=1 $ . The equation of the ellipse in the standard form, is
[IIT 1996]
Options:
A) $ \frac{{{(x-1/3)}^{2}}}{1/9}+\frac{{{(y-1)}^{2}}}{1/12}=1 $
B) $ \frac{{{(x-1/3)}^{2}}}{1/9}+\frac{{{(y+1)}^{2}}}{1/12}=1 $
C) $ \frac{{{(x-1/3)}^{2}}}{1/9}-\frac{{{(y-1)}^{2}}}{1/12}=1 $
D) $ \frac{{{(x-1/3)}^{2}}}{1/9}-\frac{{{(y+1)}^{2}}}{1/12}=1 $
Show Answer
Answer:
Correct Answer: A
Solution:
There are two common tangents to the circle $ x^{2}+y^{2}=1 $ and the hyperbola $ x^{2}-y^{2}=1. $ These are $ x=1 $ and $ x=-1 $
Out of these, $ x=1 $ is nearer to the point $ P(1/2,1) $ . Thus a directrix of the required ellipse is $ x=1. $
If $ Q(x,y) $ is any point on the ellipse, then its distance from the focus is $ QP=\sqrt{{{( x-\frac{1}{2} )}^{2}}+{{(y-1)}^{2}}} $ and its distance from the directrix $ x=1 $ is $ |x-1| $ . By definition of ellipse, $ QP=e|x-1| $
$ \Rightarrow \sqrt{{{( x-\frac{1}{2} )}^{2}}+{{(y-1)}^{2}}}=\frac{1}{2}|x-1| $
therefore $ 3x^{2}-2x+4y^{2}-8y+4=0 $ or $ \frac{{{( x-\frac{1}{3} )}^{2}}}{1/9}+\frac{{{(y-1)}^{2}}}{1/12}=1 $ .
BETA
