Coordinate Geometry Question 108
Question: A point moves such that the sum of its distances from two fixed points (ae,0) and (-ae,0) is always 2a. Then equation of its locus is
[MNR 1981]
Options:
A) $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}(1-e^{2})}=1 $
B) $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{a^{2}(1-e^{2})}=1 $
C) $ \frac{x^{2}}{a^{2}(1-e^{2})}+\frac{y^{2}}{a^{2}}=1 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ A(ae,0) $ and $ B(-ae,0) $ be two given points and $ (h,k) $ be the coordinates of the moving point P.
Now, $ PA+PB=2a $
$ \Rightarrow \sqrt{{{(h-ae)}^{2}}+k^{2}}+\sqrt{{{(h+ae)}^{2}}+k^{2}}=2a $ …..(i) But, we know that $ [{{(h-ae)}^{2}}+k^{2}]-[{{(h+ae)}^{2}}+k^{2}]=-4aeh $ …..(ii)
Dividing (ii) by (i), we get $ \sqrt{[{{(h-ae)}^{2}}+k^{2}]}-\sqrt{[{{(h+ae)}^{2}}+k^{2}]}=-2eh $ …..(iii)
Adding (i) and (iii), $ 2\sqrt{{{[h-ae)}^{2}}+k^{2}]}=2(a-eh) $ Squaring both sides, we get
$ \Rightarrow {{(h-ae)}^{2}}+k^{2}={{(a-eh)}^{2}}\Rightarrow \frac{h^{2}}{a^{2}}+\frac{k^{2}}{a^{2}(1-e^{2})}=1 $
Hence locus of P is $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}(1-e^{2})}=1 $ .
BETA
