Question:
<p>If the shortest distance of the parabola <span class="math-tex">\(y^{2}=\)</span> <span class="math-tex">\(4 x\)</span> from the centre of the circle <span class="math-tex">\(x^{2}+y^{2}-4 x-\)</span> <span class="math-tex">\(16 y+64=0\)</span> is <span class="math-tex">\(d\)</span>, then <span class="math-tex">\(d^{2}\)</span> is equal to</p>
<p style="display:inline">16</p>
<p style="display:inline">20</p>
<p style="display:inline">24</p>
<p style="display:inline">36</p>
Step-by-Step Solution
Key Concept: The shortest distance between a point and a parabola is measured along the normal to the parabola that passes through that point.
<p>Equation of circle is<br />
<span class="math-tex">$x^{2}+y^{2}-4 x-16 y+64=0$</span><br />
<span class="math-tex">$\therefore$</span> Centre <span class="math-tex">$=(2,8)$</span> and radius <span class="math-tex">$=2$</span><br />
Normal, <span class="math-tex">$y+t x=2 t+t^{2}$</span><br />
The normal will pass through <span class="math-tex">$(2,8)$</span><br />
<span class="math-tex">$\Rightarrow 8+2 t=2 t+t^{3}$</span><br />
<span class="math-tex">$\Rightarrow t^{3}=8 \Rightarrow t=2$</span><br />
<span class="math-tex">$\therefore p(t)=p\left(t^{2}, 2 t\right)=p(4,4)$</span><br />
<span class="math-tex">$\therefore$</span> Shortest distance <span class="math-tex">$=\sqrt{(2-4)^{2}+(8-4)^{2}}$</span><br />
<span class="math-tex">$=\sqrt{4+16}=\sqrt{20} \Rightarrow d^{2}=20$</span></p>
Correct Answer: B