The perimeter of a triangle is 20 and the points (-2, -3) and (-2, 3) are two of the vertices of it. Then the locus of third vertex is :
A
(x−2)249+y240=1
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B
(x+2)249+y240=1
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C
(x+2)240+y249=1
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D
(x−2)240+y249=1
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Solution
The correct option is C(x+2)240+y249=1 A(−2,−3),B(−2,3)
Let the third vertex be C(x,y)
AB + BC+CA=20
BC+CA=14
So locus of C is ellipse with A,B as focii.
2a=14
2ae=6 e=37 ∴ we get b as √40 ∴ Equation of ellipse is (x+2)240+y249=1