Coordinate Geometry Question 124
Question: The locus of a point whose difference of distance from points (3, 0) and (-3,0) is 4, is
[MP PET 2002]
Options:
A) $ \frac{x^{2}}{4}-\frac{y^{2}}{5}=1 $
B) $ \frac{x^{2}}{5}-\frac{y^{2}}{4}=1 $
C) $ \frac{x^{2}}{2}-\frac{y^{2}}{3}=1 $
D) $ \frac{x^{2}}{3}-\frac{y^{2}}{2}=1 $
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Answer:
Correct Answer: A
Solution:
Let the point be $ P(h,k). $ Given that $ PA-PB=4 $
$ \sqrt{{{(h-3)}^{2}}+k^{2}}-\sqrt{{{(h+3)}^{2}}+k^{2}}=4 $
$ \Rightarrow \sqrt{{{(h-3)}^{2}}+k^{2}}=4+\sqrt{{{(h+3)}^{2}}+k^{2}} $ Squaring both sides, we get $ {{(h-3)}^{2}}+k^{2}=16+{{(h+3)}^{2}}+k^{2}+8\sqrt{{{(h+3)}^{2}}+k^{2}} $
$ \Rightarrow h^{2}+9-6h+k^{2}=16+h^{2}+9+6h+k^{2} $
$ +8\sqrt{{{(h+3)}^{2}}+k^{2}} $
$ \Rightarrow -6h=16+6h+8\sqrt{{{(h+3)}^{2}}+k^{2}} $
$ \Rightarrow -8\sqrt{{{(h+3)}^{2}}+k^{2}}=12h+16 $ Again, squaring both sides, we get $ 64(h^{2}+9+6h+k^{2})=144h^{2}+256+2.16.12h $
$ \Rightarrow 4(h^{2}+9+6h+k^{2})=9h^{2}+16+24h $
$ \Rightarrow 4h^{2}+36+24h+4k^{2}=9h^{2}+16+24h $
$ \Rightarrow 5h^{2}-4k^{2}=20\Rightarrow \frac{h^{2}}{4}-\frac{k^{2}}{5}=1 $
Hence, the locus of point P is $ \frac{x^{2}}{4}-\frac{y^{2}}{5}=1 $ .
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