Question

If $(3m^2, -6m)$ represents the feet of the normals to the parabola $y^2 = 12x$ from $(1, 2)$, then $\sum \frac{1}{m_i}$ is-

• A. $-\frac{5}{2}$
• B. $\frac{3}{2}$
• C. 6
• D. -3

Answer 1

Answer: (A)

$-\frac{5}{2}$

Answer 2

The equation of the parabola is $y^2 = 12x$. This is in the form $y^2 = 4ax$, where $4a = 12$, so $a = 3$. The coordinates of a point on the parabola in terms of the slope $m$ of the normal at that point are $(am^2, -2am)$. Substituting $a=3$, the coordinates of the foot of the normal are $(3m^2, -6m)$. This matches the form given in the question.

The equation of the normal to the parabola $y^2 = 4ax$ at the point $(am^2, -2am)$ is given by $y = mx - 2am - am^3$. Substituting $a=3$, the equation of the normal is $y = mx - 2(3)m - 3m^3$, which simplifies to $y = mx - 6m - 3m^3$.

The normal passes through the point $(1, 2)$. Substitute $x=1$ and $y=2$ into the equation of the normal: $2 = m(1) - 6m - 3m^3$ $2 = m - 6m - 3m^3$ $2 = -5m - 3m^3$

Rearrange the equation to form a cubic equation in $m$: $3m^3 + 5m + 2 = 0$

This cubic equation gives the values of $m$ corresponding to the slopes of the normals from the point $(1, 2)$ to the parabola. Let the roots of this equation be $m_1, m_2, m_3$. These are the values of $m$ for the feet of the normals $(3m_i^2, -6m_i)$.

The question asks for the sum of the reciprocals of these roots, i.e., $\sum \frac{1}{m_i} = \frac{1}{m_1} + \frac{1}{m_2} + \frac{1}{m_3}$.

For a general cubic equation $Ax^3 + Bx^2 + Cx + D = 0$, the sum of the roots ($x_1+x_2+x_3$), the sum of the roots taken two at a time ($x_1x_2+x_2x_3+x_3x_1$), and the product of the roots ($x_1x_2x_3$) are given by Vieta's formulas:

$x_1 + x_2 + x_3 = -\frac{B}{A}$

$x_1x_2 + x_2x_3 + x_3x_1 = \frac{C}{A}$

$x_1x_2x_3 = -\frac{D}{A}$

Comparing the equation $3m^3 + 5m + 2 = 0$ with $Am^3 + Bm^2 + Cm + D = 0$, we have $A=3$, $B=0$ (since there is no $m^2$ term), $C=5$, and $D=2$.

Using Vieta's formulas for the roots $m_1, m_2, m_3$: Sum of the roots: $m_1 + m_2 + m_3 = -\frac{B}{A} = -\frac{0}{3} = 0$. Sum of the products of the roots taken two at a time: $m_1m_2 + m_2m_3 + m_3m_1 = \frac{C}{A} = \frac{5}{3}$. Product of the roots: $m_1m_2m_3 = -\frac{D}{A} = -\frac{2}{3}$.

We need to calculate $\sum \frac{1}{m_i} = \frac{1}{m_1} + \frac{1}{m_2} + \frac{1}{m_3}$. Combine the terms on the right side by finding a common denominator: $\frac{1}{m_1} + \frac{1}{m_2} + \frac{1}{m_3} = \frac{m_2m_3}{m_1m_2m_3} + \frac{m_1m_3}{m_1m_2m_3} + \frac{m_1m_2}{m_1m_2m_3} = \frac{m_1m_2 + m_2m_3 + m_3m_1}{m_1m_2m_3}$.

Substitute the values obtained from Vieta's formulas: $\sum \frac{1}{m_i} = \frac{5/3}{-2/3} = \frac{5}{3} \times \left(-\frac{3}{2}\right) = -\frac{5}{2}$.

The sum of the reciprocals of the roots $m_i$ is $-\frac{5}{2}$.