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Sum of Reciprocal Slopes of Normals to y^2 = 12x from (1, 2) | JEE Advanced Parabola

JEE Maths question with a full step-by-step solution.

Question
Given that (3mi2, −6mi)\left(3m_i^{2},\,-6m_i\right) are the feet of the normals drawn to the parabola y2=12xy^{2} = 12x from (1,2)(1,2), find the value of ∑1mi\displaystyle\sum \frac{1}{m_i}.
A−52-\dfrac52correct
B32\dfrac32
C66
D−3-3
Solution
Step 1: y2=12xy^{2}=12x is y2=4axy^{2}=4ax with a=3a=3, and the normal of slope mm is
y=mx−2am−am3=mx−6m−3m3y = mx - 2am - am^{3} = mx - 6m - 3m^{3}
 with foot (am2, −2am)=(3m2, −6m)\left(am^{2},\,-2am\right) = \left(3m^{2},\,-6m\right), which is the form given, so the mim_i are the slopes of the normals from (1,2)(1,2). Step 2: The normal passes through (1,2)(1,2):
2=m−6m−3m3 ⟹ 3m3+5m+2=02 = m - 6m - 3m^{3} \ \Longrightarrow\ 3m^{3} + 5m + 2 = 0
 Here ddm(3m3+5m+2)=9m2+5>0\dfrac{d}{dm}\left(3m^{3}+5m+2\right) = 9m^{2}+5 > 0, so the cubic is strictly increasing and only one of its roots is real; (1,2)(1,2) lies inside the parabola, since 22<12(1)2^{2} < 12(1). The relations below hold for all three roots of the cubic. Step 3: For 3m3+0⋅m2+5m+2=03m^{3} + 0\cdot m^{2} + 5m + 2 = 0,
∑mi=0,∑m1m2=53,m1m2m3=−23\sum m_i = 0, \qquad \sum m_1m_2 = \frac53, \qquad m_1m_2m_3 = -\frac23
 Step 4:
∑1mi=m2m3+m3m1+m1m2m1m2m3=5/3−2/3=−52\sum \frac{1}{m_i} = \frac{m_2m_3 + m_3m_1 + m_1m_2}{m_1m_2m_3} = \frac{5/3}{-2/3} = -\frac52
 Answer: (1).
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