Let two tangents are drawn to the curve $y^{2} - 4 (x + y) = 3 sin θ + 4 cos θ - 15, x, y, θ \in R$ from the origin whose slopes are $m_{1}, m_{2}$ . If the vertex of the curve is at maximum distance from the origin, then the value of $∣ ∣ ∣ \frac{1}{m_{1} m_{2}} ∣ ∣ ∣$ is

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Solution

Given : $y^{2} - 4 (x + y) = 3 sin θ + 4 cos θ - 15$
$\Rightarrow y^{2} - 4 y + 4 = 4 x + 3 sin θ + 4 cos θ - 11 \Rightarrow (y - 2)^{2} = 4 (x - \frac{11 - 3 sin θ - 4 cos θ}{4}) \Rightarrow (y - 2)^{2} = 4 (x - λ)$

This is a parabola whose vertex is $(λ, 2)$
Now,
$λ = \frac{11 - 3 sin θ - 4 cos θ}{4}$
We know that,
$3 sin θ + 4 cos θ \in [- 5, 5] \Rightarrow λ \in [\frac{3}{2}, 4]$
Vertex is maximum when $λ = 4$ , so the parabola becomes
$(y - 2)^{2} = 4 (x - 4)$
Now, equation of tangent of slope $m$ is
$y - 2 = m (x - 4) + \frac{1}{m}$

As the tangents are passing through origin, so putting $(0, 0)$ in the equation, we get
$- 2 = - 4 m + \frac{1}{m} \Rightarrow 4 m^{2} - 2 m - 1 = 0$
The roots of this equation are $m_{1}, m_{2}$ , then using product of roots, we get
$m_{1} m_{2} = - \frac{1}{4} ∴ ∣ ∣ ∣ \frac{1}{m_{1} m_{2}} ∣ ∣ ∣ = 4$

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