If line x+y=3 is a tangent to the ellipse with foci at (4,3) and (6,k) at point (1,2), then the value of k is
A
27
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B
17
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C
20
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D
15
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Solution
The correct option is B17 Let P(1,2) and from reflection property given tangent line is angular bisector of ∠APB So, image of B w.r.t tangent line lies on line AP
Image of (4,3) in the line x+y−3=0 is : x−41=y−31=−2(4+3−3)2=−4 ⇒x=0 and y=−1 Now points (0,−1),(1,2) and (6,k) are collinear. ⇒−1−20−1=k−26−1 ⇒k=17