Let C be a curve which is locus of the point of the intersection of lines x=2+m and my=4−m. A circle S≡(x−2)2+(y+1)2=25 intersects the curve C at four points A,B,C and D. If O is centre of the curve C, then
A
eccentricity of C is √3
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B
eccentricity of C is √2
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C
OA2+OB2+OC2+OD2=120
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D
OA2+OB2+OC2+OD2=100
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Solution
The correct option is DOA2+OB2+OC2+OD2=100 Let intersection point of the given lines be (h,k) ∴(h,k)≡(2+m,4/m−1)⇒(h−2)(k+1)=4
Hence locus will be (x−2)(y+1)=4, which represents a rectangular hyperbola centred at (2,−1)
Now given circle is (x−2)2+(y+1)2=25 ⇒r=5 ∵ Circle intersects the hyperbola at A,B,C,D
and O is centre of hyperbola as well as circle, ∴OA=OB=OC=OD=r ⇒OA2+OB2+OC2+OD2=4r2=100
and eccentricity of C is √2