Question

The normal at a point on the parabola $y^2=4x$ passes through a point P. Two more normals to this parabola also pass through P. If the centroid of the triangle formed by the feet of these three normals is G(2,0), then the abscissa of P is

• A. 4
• B. -4
• C. 5
• D. -5

Hint:

For the three normals drawn from a point $(h,k)$ to the parabola $y^2=4ax$, the sum of the slopes of the normals is zero, and the sum of the ordinates of their feet is zero. The x-coordinate of the centroid of the feet is related to $h$ by $x_G = \frac{2(h-2a)}{3}$.

Answer 1

The Correct Option is C

Step 1: Recall the properties of co-normal points.
The three points on a parabola from which normals are drawn to a common point $P(h,k)$ are called co-normal points. For a parabola $y^2=4ax$, if the feet of the three normals are $(x_1,y_1), (x_2,y_2), (x_3,y_3)$, then a key property is that the sum of their ordinates is zero: $y_1+y_2+y_3=0$.

Step 2: Find the coordinates of the centroid G.
The centroid of the triangle formed by the feet $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is: \[ G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right). \] Using the property from Step 1, the y-coordinate of the centroid is $\frac{0}{3}=0$. This matches the given centroid $G(2,0)$, confirming our understanding.

Step 3: Relate the abscissae of the feet to the coordinates of P.
Let the equation of the normal at a point $(am^2, -2am)$ be $y=mx-2am-am^3$. If this normal passes through $P(h,k)$, then $k=mh-2am-am^3$, or $am^3+(2a-h)m+k=0$. The roots $m_1,m_2,m_3$ of this cubic are the slopes of the three normals. From Vieta's formulas: $m_1+m_2+m_3=0$ and $m_1m_2+m_2m_3+m_3m_1 = \frac{2a-h}{a}$. The abscissae of the feet are $x_i = am_i^2$.

Step 4: Use the x-coordinate of the centroid.
The x-coordinate of the centroid is given as 2. \[ x_G = \frac{x_1+x_2+x_3}{3} = \frac{am_1^2+am_2^2+am_3^2}{3} = 2. \] \[ a(m_1^2+m_2^2+m_3^2) = 6. \] We use the identity $\sum m_i^2 = (\sum m_i)^2 - 2\sum m_i m_j$. \[ \sum m_i^2 = (0)^2 - 2\left(\frac{2a-h}{a}\right) = \frac{-2(2a-h)}{a} = \frac{2(h-2a)}{a}. \]

Step 5: Solve for h.
Substitute this back into the centroid equation: \[ a \left( \frac{2(h-2a)}{a} \right) = 6. \] \[ 2(h-2a) = 6 \implies h-2a = 3 \implies h = 2a+3. \] For the given parabola $y^2=4x$, we have $4a=4$, so $a=1$. \[ h = 2(1)+3 = 5. \] The abscissa of the point P is $h=5$. \[ \boxed{h=5}. \]