Question

Let the directrix of the parabola $P : y^2 = 8x$, cut $x$-axis at the point A. Let $B(\alpha, \beta), \alpha>1$, be a point on P such that the slope of AB is 3/5. If BC is a focal chord of P, then six times the area of $\Delta ABC$ is :

• A. 80
• B. 160
• C. 174
• D. 192

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We identify point A from the directrix, find point B using the given slope, determine point C as the other end of the focal chord, and then calculate the area of the triangle. 
: Key Formula or Approach: 
Parabola $y^2 = 4ax$. Directrix: $x = -a$. Focal chord ends: $(at^2, 2at)$ and $(a/t^2, -2a/t)$. 
Step 2: Detailed Explanation: 
For $y^2 = 8x$, $a = 2$. Focus $S = (2, 0)$. Directrix $x = -2$. 
Point A (intersection of directrix and x-axis) is $(-2, 0)$. 
Let $B = (2t^2, 4t)$. Slope $AB = \frac{4t - 0}{2t^2 - (-2)} = \frac{4t}{2t^2 + 2} = \frac{2t}{t^2 + 1}$. 
Given $\frac{2t}{t^2 + 1} = \frac{3}{5} \implies 10t = 3t^2 + 3 \implies 3t^2 - 10t + 3 = 0$. 
$(3t - 1)(t - 3) = 0 \implies t = 3$ or $t = 1/3$. 
If $t = 3, \alpha = 2(3^2) = 18>1$ (Accepted). 
If $t = 1/3, \alpha = 2(1/9) = 2/9 < 1$ (Rejected). 
So, $B = (18, 12)$. Since $BC$ is a focal chord, $t_C = -1/t_B = -1/3$. 
$C = (2(-1/3)^2, 4(-1/3)) = (2/9, -4/3)$. 
Area of $\Delta ABC$ with $A(-2, 0), B(18, 12), C(2/9, -4/3)$: 
Area $= \frac{1}{2} | -2(12 + 4/3) + 18(-4/3 - 0) + \frac{2}{9}(0 - 12) |$. 
Area $= \frac{1}{2} | -2(40/3) - 24 - 8/3 | = \frac{1}{2} | -80/3 - 72/3 - 8/3 |$. 
Area $= \frac{1}{2} | -160/3 | = 80/3$. 
Six times Area $= 6 \times (80/3) = 160$. 
Step 3: Final Answer: 
Six times the area of the triangle is 160.