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If the tangent to the ellipse x2 + 4y2 = 16 at the point P (θ) is a normal to the circle x2 + y2 – 8x – 4y = 0 then θ equals(a) π/2(b) π/4(c) 0(d) -π/4 |
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Answer» Correct option (a) , (c) Explanation: Given ellipse is x2/16 + y2/4 = 1 Equation of tangent at P (θ) is P(acosθ ,bsinθ) i.e. P (4cosθ , 2sinθ) is 4x cosθ/16 + 2ysin θ/4 ....(1) xcosθ + 2ysinθ = 4 .......(1) (1) is a normal to the circle x2 + y2 – 8x – 4y = The equation (1) passes through the centre (4,2) of the circle. 4cosθ + 4sinθ = 4 cosθ + sinθ = 1 squaring 1 + sin2 = 1 sin2θ = 0 2θ = 0 or π Hence, θ = 0 or π |
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