Question:
<p>If the three normals drawn to the parabola, <span class="math-tex">\(y^{2}=2 x\)</span> pass through the point <span class="math-tex">\((a, 0) a \neq 0\)</span>, then '<span class="math-tex">\(a\)</span>' must be greater than:</p>
<p style="display:inline">1</p>
<p style="display:inline"><span class="math-tex">\(-\frac{1}{2}\)</span></p>
<p style="display:inline">-1</p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{2}\)</span></p>
Step-by-Step Solution
Key Concept: The existence of three distinct normals depends on ensuring the cubic equation in slope m, derived by substituting the point (a, 0) into the general normal equation, has three unique real roots.
<p>Given,<br />
parabola is <span class="math-tex">$y^{2}=2 x$</span> ...(i)<br />
Let the equation of the normal is<br />
<span class="math-tex">$y=m x-2 a m-a m^{3}$</span> ...(ii)<br />
Using equation (i)<br />
<span class="math-tex">$4 a=2 ;$</span><br />
(Standard Equation of parabola <span class="math-tex">$y^{2}=4 a x$</span>)<br />
<span class="math-tex">$\Rightarrow a=\frac{1}{2}$</span> ...(iii)<br />
Using Equation (ii) and (iii)<br />
<span class="math-tex">$y=m x-m-\frac{1}{2} m^{3}$</span><br />
Given that normal passes through the point (<span class="math-tex">$a, 0$</span>)<br />
Hence,<br />
<span class="math-tex">$0=m(a)-m-\frac{1}{2}(m)^{3}$</span><br />
<span class="math-tex">$\Rightarrow m\left(a-1-\frac{m^{2}}{2}\right)=0$</span><br />
<span class="math-tex">$\Rightarrow m=0$</span> or <span class="math-tex">$a-1-\frac{m^{2}}{2}=0$</span><br />
<span class="math-tex">$\Rightarrow m=0$</span> or <span class="math-tex">$m^{2}=2(a-1)$</span><br />
As <span class="math-tex">$m^{2} \gt 0 \Rightarrow 2(a-1) \gt 0$</span><br />
<span class="math-tex">$a-1 \gt 0$</span><br />
<span class="math-tex">$a \gt 1$</span><br />
Hence, '<span class="math-tex">$a$</span>' must be greater than one.</p>
Correct Answer: A