Question:
<p>The shortest distance between the curves <span class="math-tex">\(y^{2}=8 x\)</span> and <span class="math-tex">\(x^{2}+y^{2}+12 y +35=0\)</span> is:</p>
<p style="display:inline"><span class="math-tex">\(\sqrt{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(3 \sqrt{2}-1\)</span></p>
<p style="display:inline"><span class="math-tex">\(2 \sqrt{3}-1\)</span></p>
<p style="display:inline"><span class="math-tex">\(2 \sqrt{2}-1\)</span></p>
Step-by-Step Solution
Key Concept: The shortest distance between a circle and another curve lies along their common normal, which necessarily passes through the center of the circle.
<p>Given parabola <span class="math-tex">$y^{2}=8 x$</span>, normal equation is:<br />
<span class="math-tex">$y=m x-4 m-2 m^{3}$</span><br />
For normal passing through <span class="math-tex">$(0,-6)$</span>:<br />
<span class="math-tex">$-6=-4 m-2 m^{3}$</span><br />
<span class="math-tex">$m^{3}+2 m-3=0$</span><br />
Real solution: <span class="math-tex">$m=1$</span><br />
Contact point <span class="math-tex">$P$</span> on parabola:<br />
<span class="math-tex">$P=\left(a m^{2},-2 a m\right)$</span><br />
<span class="math-tex">$P=\left(2(1)^{2},-4(1)\right)=(2,-4)$</span><br />
Distance from <span class="math-tex">$P$</span> to circle center <span class="math-tex">$(0,-6)$</span>:<br />
<span class="math-tex">$=\sqrt{(2-0)^{2}+(-4+6)^{2}}=2 \sqrt{2}$</span><br />
Subtract circle radius (1):<br />
Shortest distance <span class="math-tex">$=2 \sqrt{2}-1$</span></p>
Correct Answer: D