Let P ( α, β ) be a point on the parabola y ^ 2 = 4x . If P also...

Given,

Parabola ${x}^{2} = 8y$

We know that,

Chord with mid point \left({x}_{1}, {y}_{1}\right)$is$T = {S}_{1}$</p><p>$x{x}_{1} - 8\left(\dfrac{y + {y}_{1}}{2}\right) = {x}_{1} - 2 8{y}_{1}$</p><p>$⇒ x{x}_{1} - 4\left(y + {y}_{1}\right) = {x}_{1} - 2 8{y}_{1}$</p><p>Now, putting the point \left({x}_{1}, {y}_{1}\right) = \left(1, \dfrac{5}{4}\right)$ we get,

$⇒ x - 4\left(y + \dfrac{5}{4}\right) = 1 - 8 \times \dfrac{5}{4} = - 9$

$∴ x - 4y + 4 = 0 \ldots \left(i\right)$

$( \alpha, \beta )$ lies on (i) and also on ${y}^{2} = 4x$

$∴ \alpha - 4\beta + 4 = 0 \ldots \left(ii\right)</p><p>And${\beta}^{2} = 4\alpha \ldots \left(iii\right)$</p><p>Solving (ii) and (iii)</p><p>${\beta}^{2} = 4 ( 4\beta - 4 )$</p><p>$⇒ {\beta}^{2} - 16\beta + 16 = 0$</p><p>$⇒ \beta = 8 \pm 4\sqrt{3}$</p><p>And$\alpha = 4\beta - 4 = 28 \pm 16\sqrt{3}$</p><p>$∴ \left(\alpha, \beta\right) = $\left(28 + 16\sqrt{3}, 8 + 4\sqrt{3}\right)$ and ( 28 - 16\sqrt{3}, 8 - 4\sqrt{3} )$</p><p>Now, solving</p><p>$( \alpha - 28 ) ( \beta - 8 ) = ( 16\sqrt{3} ) ( 4\sqrt{3} )$</p><p>$⇒ ( \alpha - 28 ) ( \beta - 8 ) = 192$</p>$