If two tangents drawn from a point (α, β) lying on the ellipse 25 x...

Let $\alpha = \dfrac{1}{5}cos\theta, \beta = \dfrac{1}{2}sin\theta$

Equation of tangent to ${y}^{2} = 4x$ with slope $m$ will be $y = mx + \dfrac{1}{m}$

Since the tangent passes through \left(\alpha, \beta\right)$, so</p><p>$\dfrac{1}{2}sin\theta = \dfrac{m}{5}cos\theta + \dfrac{1}{m}$</p><p>$⇒ {m}^{2}\left(\dfrac{cos\theta}{5}\right) - m\left(\dfrac{1}{2}sin\theta\right) + 1 = 0$</p><p>As it is a quadratic equation in$m$so it will have two roots${m}_{1} and {m}_{2}$</p><p>Given${m}_{1} = 4{m}_{2}$</p><p>Now,${m}_{1} + {m}_{2} = \dfrac{\dfrac{1}{2}sin\theta}{\dfrac{cos\theta}{5}}$</p><p>${m}_{1}{m}_{2} = \dfrac{5}{cos\theta}$</p><p>After eliminating${m}_{1} and {m}_{2}$, we get,</p><p>$cos\theta = \dfrac{- 5 \pm \sqrt{29}}{2} ⇒ \cos^{2}\theta = \dfrac{54 \pm 10\sqrt{29}}{4}$</p><p>i.e.$\alpha = \dfrac{- 5 \pm \sqrt{29}}{10} ⇒ 10\alpha + 5 = \pm \sqrt{29}$</p><p>and${\beta}^{2} = \dfrac{1}{4}\sin^{2}\theta ⇒ 16{\beta}^{2} = - 50 \pm 10\sqrt{29}$</p><p>Hence,${\left(10\alpha + 5\right)}^{2} + {\left(16{\beta}^{2} + 50\right)}^{2} = 2929$