Question

PSQ is a focal chord of a parabola whose focus is S and vertex is A. PA and QA are produced to meet the directrix in R and T, respectively. Then $\angle RST =$

• A. $30^\circ$
• B. $60^\circ$
• C. $90^\circ$
• D. None of these

Answer 1

Answer: (C)

$90^\circ$

Answer 2

Here's a step-by-step explanation:

1. Parabola Setup:

   Consider the standard parabola $y^2 = 4ax$ with vertex $A = (0,0)$ and focus $S = (a,0)$. The directrix is the line $x = -a$.

2. Parametric Points:

   Let $P$ and $Q$ be the endpoints of the focal chord, parameterized by $t$ and $-\frac{1}{t}$ respectively. Thus:

   • $P = (at^2, 2at)$
   • $Q = \Big(a\Big(\frac{1}{t^2}\Big), -\frac{2a}{t}\Big)$

3. Lines AP and AQ:

   Find the equations of lines $AP$ and $AQ$:

   • Line $AP$: Slope $= \dfrac{2at}{at^2} = \dfrac{2}{t}$. Equation: $y = \frac{2}{t}x$. The intersection $R$ with the directrix $x = -a$ is: $$y_R = \frac{2}{t}(-a) = -\frac{2a}{t} \quad \Rightarrow \quad R = (-a, -\frac{2a}{t}).$$
   • Line $AQ$: Slope $= \dfrac{-\frac{2a}{t}}{\frac{a}{t^2}} = -2t$. Equation: $y = -2tx$. The intersection $T$ with the directrix $x = -a$ is: $$y_T = -2t(-a) = 2at \quad \Rightarrow \quad T = (-a, 2at).$$

4. Vectors SR and ST:

   Compute the vectors $\vec{SR}$ and $\vec{ST}$ with origin at $S = (a,0)$:

   • $\vec{SR} = R - S = (-a - a, -\frac{2a}{t} - 0) = (-2a, -\frac{2a}{t})$.
   • $\vec{ST} = T - S = (-a - a, 2at - 0) = (-2a, 2at)$.

5. Dot Product:

   Calculate the dot product of $\vec{SR}$ and $\vec{ST}$:

   $$\vec{SR} \cdot \vec{ST} = (-2a)(-2a) + \left(-\frac{2a}{t}\right)(2at) = 4a^2 - 4a^2 = 0.$$

   Since the dot product is zero, the angle $\angle RST$ is $90^\circ$.

Therefore, $\angle RST = 90^\circ$.