If two tangents to the ellipse x2a2+y2b2=1(a>b) make angles α and β with the major axis such that tanα+tanβ=λ, then the locus of their point of intersection is
A
x2+y2=a2
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B
x2+y2=b2
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C
x2−a2=2λxy
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D
λ(x2−a2)=2xy
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Solution
The correct option is Dλ(x2−a2)=2xy Tangent to the ellipse x2a2+y2b2=1(a>b) having slope m is y=mx±√a2m2+b2 As it passes through the point P(h,k) then ⇒k=mh±√a2m2+b2 ⇒(a2−h2)m2+2hkm+b2−k2=0
Now given tanα+tanβ=λ ⇒m1+m2=λ ⇒−2hka2−h2=λ Hence required locus is λ(x2−a2)=2xy