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Q.
If the tangent at the point (4cosθ,16√11sinθ) to the ellipse 16x2+11y2=256 is also a tangent to the circle x2+y2−2x=15, then the value of θ is
Conic Sections
Solution:
The equation of the tangent at (4cosθ,16√11sinθ) to
the ellipse 16x2+11y2=256 is 16x(4cosθ)+11y(16√11sinθ)=256
or 4xcosθ+√11ysinθ=16
This touches the circle (x−1)2+y2=42 if |4cosθ−16√16cos2θ+11sin2θ|=4 ⇒(cosθ−4)2=16cos2θ+11sin2θ ⇒15cos2θ+11sin2θ+8cosθ−16=0 ⇒4cos2θ+8cosθ−5=0 ⇒(2cosθ−1)(2cosθ+5)=0 ⇒cosθ=12 ⇒θ=±π3