Let $A(4,-4)$ and $B(9,6)$ be points on the parabola, $y^2 + 4x$. Let $C$ be chosen on the arc $AOB$ of the parabola, where $O$ is the origin, such that the area of $\Delta ACB$ is maximum. Then, the area (in s units) of $\Delta ACB$, is :

Question

Let $A$ be the focus of the parabola $y^2=8x$. Let the line $y=mx+c$ intersect the parabola at two distinct points $B$ and $C$. If the centroid of triangle $ABC$ is $\left(\frac{7}{3},\frac{4}{3}\right)$, then $(BC)^2$ is equal to:

Answer 1

To solve this problem, we need to find the maximum area of triangle $ \Delta ACB $ where points $ A(4, -4) $ and $ B(9, 6) $ are on the parabola $ y^2 = 4x $, and $ C $ is a point on the arc $ AOB $ such that $ O $ is the origin.

Step 1: Parametrize the Parabola
The equation of the parabola is $ y^2 = 4x $. This can be parametrized using $ t $ as $ x = t^2 $ and $ y = 2t $. The coordinates of a point on the parabola in terms of parameter $ t $ are $ (t^2, 2t) $.

Step 2: Calculate the Points A and B
For point $ A(4, -4) $:

We equate $ 2t = -4 $ giving $ t = -2 $.
This makes $ t^2 = 4 $ which matches the $ x $-coordinate of $ A $.

For point $ B(9, 6) $:

We equate $ 2t = 6 $ giving $ t = 3 $.
This makes $ t^2 = 9 $ which matches the $ x $-coordinate of $ B $.

Step 3: Find the Maximum Area of Triangle $ \Delta ACB $
To find the area of the triangle $ \Delta ACB $ given vertices $ A(4, -4) $, $ B(9, 6) $, and $ C(t_1^2, 2t_1) $:

The area $ \Delta $ of triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is given by:

$\Delta = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$

Substitute $ x_1 = 4 $, $ y_1 = -4 $, $ x_2 = t_1^2 $, $ y_2 = 2t_1 $, $ x_3 = 9 $, $ y_3 = 6 $:

$\Delta = \frac{1}{2} \left| 4(2t_1 - 6) + t_1^2(6 + 4) + 9(-4 - 2t_1) \right|$

Calculate further:

$ = \frac{1}{2} \left| 8t_1 - 24 + 10t_1^2 - 36 - 18t_1 \right| $
$ = \frac{1}{2} \left| 10t_1^2 - 10t_1 - 60 \right| $

The area is maximum when $-b/2a = t_1$. Replace $ a = 10 $, $ b = -10 $ to find $ t_1 $:

$ t_1 = \frac{-(-10)}{2 \times 10} = \frac{1}{2} $

The corresponding coordinate $ C = \left(\left( \frac{1}{2} \right)^2, 2\left(\frac{1}{2}\right)\right) = \left(\frac{1}{4}, 1\right) $. Substituting back yields maximum area of $\Delta ACB = \frac{125}{4} = 31 \frac{1}{4} $.

Conclusion:
The maximum area of $ \Delta ACB $ is $ 31 \frac{1}{4} $, which matches with the given option.

Answer 2