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Question: Let a tangent to ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ meets its director circle at P and Q. I...

Let a tangent to ellipse x29+y24=1\frac{x^2}{9} + \frac{y^2}{4} = 1 meets its director circle at P and Q. If λ\lambda denotes the product of slopes of CP and CQ where C is centre of ellipse then λ\lambda is greater than

A

94\frac{9}{4}

B

49\frac{4}{9}

C

−12-\frac{1}{2}

D

−94-\frac{9}{4}

Answer

C and D

Explanation

Solution

The equation of the ellipse is x29+y24=1\frac{x^2}{9} + \frac{y^2}{4} = 1, so a2=9a^2=9 and b2=4b^2=4. The center of the ellipse is C=(0,0)C=(0,0). The director circle of the ellipse is x2+y2=a2+b2=9+4=13x^2 + y^2 = a^2 + b^2 = 9 + 4 = 13. Let the tangent to the ellipse be y=mx±a2m2+b2y = mx \pm \sqrt{a^2m^2 + b^2}. The points P and Q are the intersections of the tangent and the director circle. The equation of the lines CP and CQ passing through the origin and points P and Q can be obtained by homogenizing the equation of the director circle with the tangent equation. The equation of the tangent is xacos⁡θ+ybsin⁡θ=1\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1. Substituting 1=x3cos⁡θ+y2sin⁡θ1 = \frac{x}{3}\cos\theta + \frac{y}{2}\sin\theta into x2+y2=13x^2+y^2=13: x2+y2=13(x3cos⁡θ+y2sin⁡θ)2x^2 + y^2 = 13 \left(\frac{x}{3}\cos\theta + \frac{y}{2}\sin\theta\right)^2 x2+y2=13(x29cos⁡2θ+y24sin⁡2θ+xy3cos⁡θsin⁡θ)x^2 + y^2 = 13 \left(\frac{x^2}{9}\cos^2\theta + \frac{y^2}{4}\sin^2\theta + \frac{xy}{3}\cos\theta\sin\theta\right) Rearranging this into a homogeneous quadratic equation Ax2+2Hxy+By2=0Ax^2 + 2Hxy + By^2 = 0: x2(1−139cos⁡2θ)+y2(1−134sin⁡2θ)−133xycos⁡θsin⁡θ=0x^2\left(1 - \frac{13}{9}\cos^2\theta\right) + y^2\left(1 - \frac{13}{4}\sin^2\theta\right) - \frac{13}{3}xy\cos\theta\sin\theta = 0 The product of the slopes of CP and CQ (which are the lines represented by this homogeneous equation) is λ=AB\lambda = \frac{A}{B}. λ=1−139cos⁡2θ1−134sin⁡2θ=1−139(1−sin⁡2θ)1−134sin⁡2θ=−49+139sin⁡2θ1−134sin⁡2θ\lambda = \frac{1 - \frac{13}{9}\cos^2\theta}{1 - \frac{13}{4}\sin^2\theta} = \frac{1 - \frac{13}{9}(1-\sin^2\theta)}{1 - \frac{13}{4}\sin^2\theta} = \frac{-\frac{4}{9} + \frac{13}{9}\sin^2\theta}{1 - \frac{13}{4}\sin^2\theta} λ=13sin⁡2θ−494−13sin⁡2θ4=49⋅13sin⁡2θ−44−13sin⁡2θ=49⋅−(4−13sin⁡2θ)4−13sin⁡2θ=−49\lambda = \frac{\frac{13\sin^2\theta - 4}{9}}{\frac{4 - 13\sin^2\theta}{4}} = \frac{4}{9} \cdot \frac{13\sin^2\theta - 4}{4 - 13\sin^2\theta} = \frac{4}{9} \cdot \frac{-(4 - 13\sin^2\theta)}{4 - 13\sin^2\theta} = -\frac{4}{9}. We are given that λ\lambda is greater than one of the options. We have λ=−49≈−0.444\lambda = -\frac{4}{9} \approx -0.444. Checking the options: (A) −49>94-\frac{4}{9} > \frac{9}{4} is false. (B) −49>49-\frac{4}{9} > \frac{4}{9} is false. (C) −49>−12-\frac{4}{9} > -\frac{1}{2} is true, since −0.444>−0.5-0.444 > -0.5. (D) −49>−94-\frac{4}{9} > -\frac{9}{4} is true, since −0.444>−2.25-0.444 > -2.25. Both options (C) and (D) satisfy the condition λ>option\lambda > \text{option}.