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Locus of a Point Dividing an Ellipse Tangent Intercept in Ratio 2 : 1 | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
A tangent PQPQ to the ellipse x2a2+y2b2=1\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1 cuts the positive xx and yy axes at PP and QQ respectively. The point RR divides PQPQ internally in the ratio 2:12 : 1. Given that the locus of RR is a2x2+4b2y2=λ\dfrac{a^{2}}{x^{2}} + \dfrac{4b^{2}}{y^{2}} = \lambda, find λ\lambda.
Solution
Answer: 9 (± 0.01)
Step 1: At the point (acos⁡θ, bsin⁡θ)(a\cos\theta,\ b\sin\theta) the tangent is
xcos⁡θa+ysin⁡θb=1\frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1
 Both intercepts are positive, so cos⁡θ>0\cos\theta > 0 and sin⁡θ>0\sin\theta > 0. Step 2:
y=0⇒P(acos⁡θ, 0),x=0⇒Q(0, bsin⁡θ)y = 0 \Rightarrow P\left(\frac{a}{\cos\theta},\ 0\right), \qquad x = 0 \Rightarrow Q\left(0,\ \frac{b}{\sin\theta}\right)
 Step 3: PR:RQ=2:1PR : RQ = 2 : 1, so by the section formula
R=1⋅P+2⋅Q3R = \frac{1\cdot P + 2\cdot Q}{3}
 and writing R=(h,k)R = (h,k),
h=13⋅acos⁡θ=a3cos⁡θ,k=23⋅bsin⁡θ=2b3sin⁡θh = \frac{1}{3}\cdot\frac{a}{\cos\theta} = \frac{a}{3\cos\theta}, \qquad k = \frac{2}{3}\cdot\frac{b}{\sin\theta} = \frac{2b}{3\sin\theta}
 Step 4:
cos⁡θ=a3h,sin⁡θ=2b3k\cos\theta = \frac{a}{3h}, \qquad \sin\theta = \frac{2b}{3k}
 Step 5: cos⁡2θ+sin⁡2θ=1\cos^{2}\theta + \sin^{2}\theta = 1 gives
a29h2+4b29k2=1 ⟹ a2h2+4b2k2=9\frac{a^{2}}{9h^{2}} + \frac{4b^{2}}{9k^{2}} = 1 \ \Longrightarrow\ \frac{a^{2}}{h^{2}} + \frac{4b^{2}}{k^{2}} = 9
 Step 6: Replacing (h,k)(h,k) by (x,y)(x,y),
a2x2+4b2y2=9 ⟹ λ=9\frac{a^{2}}{x^{2}} + \frac{4b^{2}}{y^{2}} = 9 \ \Longrightarrow\ \lambda = 9
 Answer: 99.
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