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Angle Between Tangents at the Ends of a Chord of Contact | JEE Advanced Ellipse

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

Question
The line x2cos⁡θ+y1sin⁡θ=1\dfrac{x}{2}\cos\theta + \dfrac{y}{1}\sin\theta = 1, with θ∈(π18,π15)\theta \in \left(\dfrac{\pi}{18},\dfrac{\pi}{15}\right), meets the ellipse x2+2y2=6x^{2}+2y^{2}=6 at PP and QQ. Find the angle between the tangents to the ellipse at PP and QQ.
Aπ16\dfrac{\pi}{16}
Bπ6\dfrac{\pi}{6}
Cπ8\dfrac{\pi}{8}
Dπ2\dfrac{\pi}{2}correct
Solution
Step 1: The tangents at PP and QQ meet at some point R(h,k)R(h,k), so the line PQPQ is the chord of contact of RR with respect to the ellipse. Step 2: For x2+2y2=6x^{2} + 2y^{2} = 6 the chord of contact of R(h,k)R(h,k) is T=0T = 0:
hx+2ky=6hx + 2ky = 6
 Step 3: Multiplying x2cos⁡θ+ysin⁡θ=1\dfrac{x}{2}\cos\theta + y\sin\theta = 1 by 66,
3xcos⁡θ+6ysin⁡θ=63x\cos\theta + 6y\sin\theta = 6
 so comparing coefficients,
h=3cos⁡θ,2k=6sin⁡θ ⟹ k=3sin⁡θh = 3\cos\theta, \qquad 2k = 6\sin\theta \ \Longrightarrow\ k = 3\sin\theta
 Step 4:
h2+k2=9(cos⁡2θ+sin⁡2θ)=9h^{2} + k^{2} = 9\left(\cos^{2}\theta + \sin^{2}\theta\right) = 9
 So RR lies on the circle x2+y2=9x^{2}+y^{2} = 9. Step 5: Writing the ellipse as x26+y23=1\dfrac{x^{2}}{6} + \dfrac{y^{2}}{3} = 1, its director circle is
x2+y2=a2+b2=6+3=9x^{2}+y^{2} = a^{2}+b^{2} = 6 + 3 = 9
 which is exactly the circle found in Step 4. Every point of the ellipse has x2+y2=6−y2≤6<9x^{2}+y^{2} = 6 - y^{2} \le 6 < 9, so RR is outside the ellipse and the two tangents are real. Step 6: Tangents drawn from any point of the director circle are perpendicular, so the angle between the tangents at PP and QQ is
π2\frac{\pi}{2}
 The restriction on θ\theta only makes PP and QQ two distinct real points; it does not affect the angle. Answer: (4).
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