An isosceles $Δ A B C$ is inscribed in the circle $x^{2} + y^{2} = 16, A = (0, 4) .$ From points A, B and C three ordinates are drawn which cut the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ at the points P, Q and R respectively such that A and P are on the opposite sides, Q and B are on the same side and C and R are on the same side of the major axis. If $Δ P Q R$ is right angled at P and $θ$ is the smallest angle of the $Δ A B C,$ then find the value of $16 {tan}^{2} θ + 8 tan θ + 9 .$

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answer is 24.

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Detailed Solution

$Q \equiv (- 4 cos ϕ, 3 sin ϕ), R \equiv (+ 4 cos ϕ, 3 sin ϕ) (0, 4)$

$P Q ⊥ P R$

$\begin{array}{r} \Rightarrow & \frac{3 (sin ϕ + 1)}{4 cos ϕ} \frac{3 (sin ϕ + 1)}{- 4 cos ϕ} = - 1 \\ \Rightarrow & 25 {sin}^{2} ϕ + 18 sin ϕ - 7 = 0 \\ \Rightarrow & sin ϕ = \frac{7}{25} (sin ϕ = - 1 r e j e c t e d) \end{array}$

$\begin{array}{r} tan \frac{ϕ}{2} = \sqrt{(1 - \frac{24}{25}) / (1 + \frac{24}{25})} = \frac{1}{7} \\ \Rightarrow & θ = ∠ A B C = \frac{1}{2} (\frac{π}{2} - ϕ) \\ tan θ = \frac{1 - tan \frac{ϕ}{2}}{1 + tan \frac{ϕ}{2}} = \frac{1 - \frac{1}{7}}{1 + \frac{1}{7}} = \frac{3}{4} \\ \Rightarrow & 16 {tan}^{2} θ + 8 tan θ + 9 \\ = 16 \times \frac{9}{16} + 8 \times \frac{3}{4} + 9 = 24 \end{array}$