Parabola
Conormal Points and Normals
Grade 11
Question:
<p>Find the locus of the point through which pass three normals to the parabola <i>y</i><sup>2</sup> = 4<i>ax</i> such that two of them make angles α and β respectively with the axis such that tan α tan β = 2.</p>
<p>(a) <i>x</i><sup>2</sup> - 4<i>ay</i> = 0</p>
<p>(b) <i>y</i><sup>2</sup> - 4<i>ax</i> = 0</p>
<p>(c) <i>x</i><sup>2</sup> + 4<i>ay</i> = 0</p>
<p>(d) <i>y</i><sup>2</sup> + 4<i>ax</i> = 0</p>
Step-by-Step Solution
Key Concept: Three normals pass through a point if the slopes satisfy a cubic equation; use Vieta's formulas and the given constraint on the product of two slopes.
<p><strong>Solution:</strong></p><p>Let (<i>h</i>, <i>k</i>) be the point of intersection of three normals to the parabola <i>y</i><sup>2</sup> = 4<i>ax</i>.</p><p>The equation of any normal to <i>y</i><sup>2</sup> = 4<i>ax</i> is: $y = mx - 2am - am^3$</p><p>If it passes through (<i>h</i>, <i>k</i>), then: $k = mh - 2am - am^3$</p><p>$\Rightarrow am^3 + m(2a - h) + k = 0$ ... (i)</p><p>Let roots of Eq. (i) be <i>m</i><sub>1</sub>, <i>m</i><sub>2</sub>, <i>m</i><sub>3</sub>.</p><p>From Eq. (i): $m_1 m_2 m_3 = -\frac{k}{a}$ ... (ii)</p><p>Given: <i>m</i><sub>1</sub> = tan α, <i>m</i><sub>2</sub> = tan β and tan α tan β = 2 ... (iii)</p><p>Therefore: $m_1 m_2 = 2$</p><p>From Eqs. (ii) and (iii): $2m_3 = -\frac{k}{a}$</p><p>The locus is obtained by substituting <i>h</i> → <i>x</i>, <i>k</i> → <i>y</i>:</p><p>$y^2 - 4ax = 0$</p>
Correct Answer: b