<p>The point at which the line y = mx + c touches the parabola y<sup>2</sup> = 4ax is</p>
<p style="display:inline"><span class="math-tex">\(\left(-\frac{a}{m^{2}}, \frac{2 a}{m}\right)\)</span></p>
<p style="display:inline"><span class="math-tex">\(\left(\frac{a}{m^{2}}, \frac{-2 a}{m^{2}}\right)\)</span></p>
<p style="display:inline"><span class="math-tex">\(\left(\frac{a}{m^{2}}, \frac{2 a}{m}\right)\)</span></p>
<p style="display:inline"><span class="math-tex">\(\left(-\frac{a}{m^{2}},-\frac{2 a}{m^{2}}\right) \)</span></p>
Step-by-Step Solution
Key Concept: The point of tangency is found by setting the discriminant of the quadratic equation formed by the intersection of the line and the parabola to zero.
<p>The line y = mx + c is tangent to the parabola<br />
y<sup>2</sup> = 4ar<br />
<span class="math-tex">$\Rightarrow$</span> m<sup>2</sup>x<sup>2</sup> + (2mc - 4a)x + c<sup>2</sup> = 0 ...(i)<br />
has equal roots<br />
<span class="math-tex">$\Rightarrow$</span> discriminant = 0<br />
<span class="math-tex">$\Rightarrow$</span> a = mc ...(ii)<br />
Substitute in (i) to get (mx - c)<sup>2</sup> = 0<br />
<span class="math-tex">$\Rightarrow x=\frac{c}{m}=\frac{a}{m^{2}}$</span> ...[From (ii)]<br />
Substitute in y<sup>2</sup> = 4ax to get y = <span class="math-tex">$\frac{2 a}{m}$</span></p>
Correct Answer: C