Question: The locus of a point whose difference of distance from points (3, 0) and $ (-3,0) $ is 4, is
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 $ (x,y). $
Then. $ \sqrt{{{(x-3)}^{2}}+y^{2}}-\sqrt{{{(x+3)}^{2}}+y^{2}}=4 $
$ \Rightarrow $ $ \sqrt{{{(x-3)}^{2}}+y^{2}}=4+\sqrt{{{(x+3)}^{2}}+y^{2}} $
On squaring both sides, we get
$ {{(x-3)}^{2}}+y^{2}=16+{{(x+3)}^{2}}+y^{2}+8\sqrt{{{(x+3)}^{2}}+y^{2}} $
$ \Rightarrow $ $ -6x-6x-16=8\sqrt{{{(x+3)}^{2}}+y^{2}} $
$ \Rightarrow $ $ -12x-16=8\sqrt{{{(x+3)}^{2}}+y^{2}} $
$ \Rightarrow $ $ -3x-4=2\sqrt{{{(x+3)}^{2}}+y^{2}} $
Again, on squaring both sides, we get
$ 9x^{2}+16+24x=4,[{{(x+3)}^{2}}+y^{2}] $
$ \Rightarrow $ $ 9x^{2}+16+24x=4,[x^{2}+9+6x+y^{2}] $
$ \Rightarrow $ $ 9x^{2}+16+24x=4x^{2}+36+24x+4y^{2} $
$ \Rightarrow $ $ 5x^{2}-4y^{2}=20 $
$ \Rightarrow $ $ \frac{x^{2}}{4}-\frac{y^{2}}{5}=1 $
BETA
