If $\mathrm{P}(\mathrm{h}, \mathrm{k})$ be a point on the parabola $x=4 y^{2}$, which is nearest to the point $\mathrm{Q}(0,33)$, then the distance of $\mathrm{P}$ from the directrix of the parabola $\quad y^{2}=4(x+y)$ is equal to :
Solution
<p>Equation of normal</p>
<p>$y = - tx + 2.{1 \over {16}}t + {1 \over {16}}{t^3}$</p>
<p>$33 = {t \over 8} + {{{t^3}} \over {16}}$</p>
<p>$\Rightarrow {t^3} + 2t = 528$</p>
<p>$t = 8$</p>
<p>$(a{t^2},2at) = (4,1)$</p>
<p>Distance from $x = - 2$</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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