Question:
<p>The slope of the line touching both the parabolas y<sup>2</sup> = 4x and x<sup>2</sup> = - 32y is</p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{3}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{8}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{2}{3}\)</span></p>
Step-by-Step Solution
Key Concept: Find the common tangent by substituting the slope-form tangent equation of one parabola into the other parabola and solving for the slope using the tangency condition where the discriminant of the resulting quadratic equals zero.
<p>Let the tangent to parabola be y = mx + a/m, if it touches the other curve, then D = 0, to get the value of m.<br />
For parabola, y<sup>2</sup> = 4x<br />
Let <span class="math-tex">$y=m x+\frac{1}{m}$</span> be tangent line and it touches the parabola x<sup>2</sup> = -32y<br />
<span class="math-tex">$\therefore \quad x^{2}=-32\left(m x+\frac{1}{m}\right)$</span><br />
<span class="math-tex">$\Rightarrow \quad x^{2}+32 m x+\frac{32}{m}=0$</span><br />
D = 0<br />
<span class="math-tex">$\because \quad(32 m)^{2}-4 \cdot\left(\frac{32}{m}\right)=0 \Rightarrow m^{3}=1 / 8$</span><br />
<span class="math-tex">$\therefore \quad m=1 / 2$</span></p>
Correct Answer: A