Question

If the normals drawn at the points $P\left(\frac{3}{4}, \frac{3}{2}\right)$ and $Q(3,3)$ on the parabola $y^2 = 3x$ intersect again on $y^2=3x$ at R, then R =

• A. $(12,6)$
• B. $\left(\frac{27}{4}, \frac{9}{2}\right)$
• C. $\left(\frac{3}{16}, \frac{3}{4}\right)$
• D. $\left(\frac{1}{12}, \frac{1}{2}\right)$

Hint:

For the parabola $y^2=4ax$, remember these key properties of normals: 1. If the normal at $t_1$ meets the parabola again at $t_2$, then $t_2 = -t_1 - \frac{2}{t_1}$. 2. If the normals at $t_1$ and $t_2$ meet on the parabola at $t_3$, then $t_1+t_2+t_3=0$ is NOT the property. The correct property is $t_3=-(t_1+t_2)$ under the condition $t_1 t_2 = 2$. 3. If normals at three points $t_1, t_2, t_3$ are concurrent, then $t_1+t_2+t_3=0$ and the sum of their ordinates is zero. In this problem, the condition $t_P t_Q = 1 \times 2 = 2$ is satisfied, confirming the applicability of the rule $t_R = -(t_P+t_Q)$.

Answer 1

The Correct Option is B

Step 1: Identify Parabola Parameters: Given parabola \( y^2 = 3x \). Comparing with \( y^2 = 4ax \), we get \( 4a = 3 \implies a = \frac{3}{4} \). Parametric coordinates are \( (at^2, 2at) \).
Step 2: Find Parameters for P and Q: For P \( \left(\frac{3}{4}, \frac{3}{2}\right) \): \( 2at_1 = \frac{3}{2} \implies 2\left(\frac{3}{4}\right)t_1 = \frac{3}{2} \implies \frac{3}{2}t_1 = \frac{3}{2} \implies t_1 = 1 \). For Q \( (3,3) \): \( 2at_2 = 3 \implies 2\left(\frac{3}{4}\right)t_2 = 3 \implies \frac{3}{2}t_2 = 3 \implies t_2 = 2 \).
Step 3: Find Intersection of Normals: The point of intersection of normals at \( t_1 \) and \( t_2 \) is given by: \[ x = 2a + a(t_1^2 + t_1t_2 + t_2^2) \] \[ y = -at_1t_2(t_1+t_2) \] Note: The question states they intersect "again on \( y^2=3x \) at R". This implies the intersection point R lies on the parabola. We calculate the intersection coordinates directly. Substituting \( a = \frac{3}{4}, t_1 = 1, t_2 = 2 \): \[ x = 2\left(\frac{3}{4}\right) + \frac{3}{4}(1^2 + 1(2) + 2^2) = \frac{3}{2} + \frac{3}{4}(1+2+4) = \frac{3}{2} + \frac{21}{4} = \frac{6+21}{4} = \frac{27}{4} \] \[ y = -\frac{3}{4}(1)(2)(1+2) = -\frac{3}{4}(2)(3) = -\frac{18}{4} = -\frac{9}{2} \]
Step 4: Verify R is on Parabola: Check if \( R\left(\frac{27}{4}, -\frac{9}{2}\right) \) satisfies \( y^2 = 3x \). LHS: \( \left(-\frac{9}{2}\right)^2 = \frac{81}{4} \). RHS: \( 3\left(\frac{27}{4}\right) = \frac{81}{4} \). It matches. Thus, \( R = \left(\frac{27}{4}, -\frac{9}{2}\right) \).