Let P be a point on the parabola x ^ 2 = 4y . If the distance of P...

Let $P\left(2t, {t}^{2}\right).</p><p>Equation normal at$P to {x}^{2} = 4y$be</p><p>$y - {t}^{2} = - $\dfrac{1}{t}\left(x - 2t\right).</p><p>It passes a normal through$\left(- 3, 0\right).

$⇒ 0 - {t}^{2} = -$\dfrac{1}{t}\left(- 3 - 2t\right)

$⇒ {t}^{3} + 2t + 3 = 0$

$⇒ \left(t + 1\right)\left({t}^{2} - t + 3\right)$ = 0$</p><p>$⇒ t = - 1$</p><p>So, point$P$is \left(- 2, 1\right).</p><p>Equation$ of a tangent to{x}^{2} = 4y$at \left(- 2, 1\right).</p><p>$⇒ x \left(- 2\right)$= 2 \left(y + 1\right)$

$∴ x + y + 1 = 0$.