Match Elements of List I with elements of List II.
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Solution
A: Tangent to ellipse at P(ϕ) is x4cosϕ+y2sinϕ=1. It must pass through the centre of the circle. Hence, 44cosϕ+22sinϕ=1 ⇒cosϕ+sinϕ=1
√1−sin2ϕ+sinϕ=1
1−sin2ϕ=(1−sinϕ)2
sinϕ−sin2ϕ=0 ⇒1+sin2ϕ=1 ⇒ϕ=0 or π2 ⇒ϕ2=0 or π4 B: Consider any point P (√6cosθ,√2sinθ) on ellipse x26+y22=1 Given that OP=2 ⇒6cos2θ+2sin2θ=4 ⇒4cos2θ=2 ⇒cosθ=±1√2 ⇒θ=π4 or 5π4 C: Solving the equation of ellipse and parabola (eliminating x2), we have ⇒y−1+4y2=4 ⇒4y2+y−5=0 ⇒4y2+y−5=0 ⇒(4y+5)(y−1)=0 →y=1,x=0 The curves touch at (0,1). So, the angle of intersection is 0. D: The normal at P(acosθ,sinθ) is axcosθ−bxsinθ=a2−b2 where a2=14,b2=5. It meets the curve again at Q(2θ), i.e. (acos2θ,bsin2θ).