Consider the parabola y ^ 2 = 4x . Let S be the focus of the...

On plotting the diagram of the given value we have,

Now let ${P}_{1}\left({t}^{2}, 2t\right)$ then tangent at ${P}_{1}$ will be $ty = x + {t}^{2}$

Since it passes through \left(- 2, 1\right)$</p><p>$⇒ {t}^{2} - t - 2 = 0$</p><p>$⇒ t = 2, - 1$</p><p>So, we get points${P}_{1}\left(4, 4\right)$and${P}_{2}\left(1, - 2\right)$</p><p>Now finding the equation of$S{P}_{1}: 4x - 3y - 4 = 0$</p><p>And equation of$S{P}_{2}: x - 1 = 0$</p><p>Now finding the point by foot of point on line formula,</p><p>${Q}_{1}: \dfrac{{x}_{1} + 2}{4} = \dfrac{{y}_{1} - 1}{- 3} = \dfrac{- \left(- 8 - 3 - 4\right)}{25} = \dfrac{3}{5}$</p><p>We get${x}_{1} = \dfrac{2}{5}, {y}_{1} = \dfrac{- 4}{5}$</p><p>Similarly for${Q}_{2} = \left(1, 1\right)$</p><p>Now using the distance formula we get,$S{Q}_{1} = \sqrt{{\left(1 - \dfrac{2}{5}\right)}^{2} + {\left(\dfrac{4}{5}\right)}^{2}} = 1$</p><p>${Q}_{1}{Q}_{2} = \sqrt{\dfrac{9}{25} + \dfrac{81}{25}} = \sqrt{\dfrac{90}{25}} = \dfrac{3\sqrt{10}}{5}$</p><p>$P{Q}_{1} = \sqrt{\dfrac{144}{25} + \dfrac{81}{25}} = 3$</p><p>$S{Q}_{2} = 1.