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: Understanding the Concept
This problem involves the properties of normals to a parabola. A standard result states that if the normals at two points with parameters $t_1$ and $t_2$ on a parabola meet on the parabola again at a third point with parameter $t_3$, then $t_3 = -(t_1+t_2)$. Also, this condition holds if $t_1t_2=2$.
Step 2: Key Formula or Approach
1. Identify the parameter $a$ for the parabola $y^2=3x$. 2. Find the parameters $t_P$ and $t_Q$ corresponding to the points P and Q using the parametric form of the parabola $(at^2, 2at)$. 3. Use the property $t_R = -(t_P+t_Q)$ to find the parameter for the point R. 4. Use the parameter $t_R$ to find the coordinates of R.
Step 3: Detailed Explanation
1. Identify parabola parameter: The equation of the parabola is $y^2=3x$. Comparing this with the standard form $y^2=4ax$, we have $4a=3$, which gives $a = \frac{3}{4}$. 2. Find parameters for P and Q: The parametric coordinates of a point on this parabola are $(at^2, 2at) = \left(\frac{3}{4}t^2, \frac{3}{2}t\right)$. For point $P\left(\frac{3}{4}, \frac{3}{2}\right)$: The y-coordinate is $2at_P = \frac{3}{2}t_P = \frac{3}{2} \implies t_P = 1$. Let's check the x-coordinate: $at_P^2 = \frac{3}{4}(1)^2 = \frac{3}{4}$. This matches. So, $t_P = 1$. For point $Q(3,3)$: The y-coordinate is $2at_Q = \frac{3}{2}t_Q = 3 \implies t_Q = 2$. Let's check the x-coordinate: $at_Q^2 = \frac{3}{4}(2)^2 = \frac{3}{4}(4) = 3$. This matches. So, $t_Q = 2$. 3. Find the parameter for R: The parameter of the point of intersection R, $t_R$, is given by the relation: \[ t_R = -(t_P + t_Q) \] \[ t_R = -(1 + 2) = -3 \] 4. Find the coordinates of R: Now we find the coordinates of R using its parameter $t_R = -3$. x-coordinate of R: $x_R = at_R^2 = \frac{3}{4}(-3)^2 = \frac{3}{4}(9) = \frac{27}{4}$. y-coordinate of R: $y_R = 2at_R = \frac{3}{2}(-3) = -\frac{9}{2}$. So the coordinates of R are $\left(\frac{27}{4}, -\frac{9}{2}\right)$. Note on the options: The calculated y-coordinate is $-\frac{9}{2}$. Option (B) has the correct x-coordinate and the correct magnitude for the y-coordinate, but with a positive sign. This is a common type of error in exam questions or answer keys. Based on the derived result, the correct point is $\left(\frac{27}{4}, -\frac{9}{2}\right)$, and we select the closest option. Step 4: Final Answer
The coordinates of R are calculated to be $\left(\frac{27}{4}, -\frac{9}{2}\right)$. Option (B) is $\left(\frac{27}{4}, \frac{9}{2}\right)$, which has the same components but a different sign for the y-coordinate. Assuming a typo in the option, (B) is the intended answer.