Question

Let a focus of the ellipse E: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be S(4, 0) and its eccentricity be $\frac{4}{5}$. If the point P(3, $\alpha$) lies on E and O is the origin, then the area of $\Delta$POS is equal to:

• A. 12/5
• B. 14/5
• C. 24/5
• D. 48/5

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given information about an ellipse (a focus and eccentricity) and a point that lies on it. We need to find the area of a triangle formed by this point, the given focus, and the origin. 
Step 2: Key Formula or Approach: 
1. The standard equation for an ellipse centered at the origin with foci on the x-axis is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (with $a>b$). 
2. The coordinates of the foci are $(\pm ae, 0)$. 
3. The relationship between $a, b,$ and $e$ is $b^2 = a^2(1-e^2)$. 
4. The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$. A simpler method is (1/2) * base * height if the base is along an axis. 
Step 3: Detailed Explanation: 
First, let's find the parameters $a$ and $b$ of the ellipse. 
The focus is given as S(4, 0). For a standard ellipse, the foci are at $(\pm ae, 0)$. 
So, $ae = 4$. 
The eccentricity is given as $e = \frac{4}{5}$. 
Substituting the value of $e$: 
$a \left(\frac{4}{5}\right) = 4 \implies a = 5$. 
Now, find $b^2$ using the relation $b^2 = a^2(1-e^2)$. 
$b^2 = 5^2 \left(1 - \left(\frac{4}{5}\right)^2\right) = 25 \left(1 - \frac{16}{25}\right) = 25 \left(\frac{25-16}{25}\right) = 25 \left(\frac{9}{25}\right) = 9$. 
So, the equation of the ellipse E is: 
\[ \frac{x^2}{5^2} + \frac{y^2}{3^2} = 1 \implies \frac{x^2}{25} + \frac{y^2}{9} = 1 \] The point P(3, $\alpha$) lies on the ellipse. So, its coordinates must satisfy the ellipse's equation. 
Substitute $x=3$ and $y=\alpha$: 
\[ \frac{3^2}{25} + \frac{\alpha^2}{9} = 1 \] \[ \frac{9}{25} + \frac{\alpha^2}{9} = 1 \] \[ \frac{\alpha^2}{9} = 1 - \frac{9}{25} = \frac{16}{25} \] \[ \alpha^2 = 9 \times \frac{16}{25} = \frac{144}{25} \] \[ \alpha = \pm \sqrt{\frac{144}{25}} = \pm \frac{12}{5} \] Now we need to find the area of the triangle POS. The vertices are: 
P = $(3, \alpha) = (3, \pm 12/5)$ 
O = $(0, 0)$ 
S = $(4, 0)$ 
We can use the formula Area = $\frac{1}{2} \times \text{base} \times \text{height}$. 
Let's take the segment OS as the base of the triangle. This segment lies on the x-axis. 
Base length = Distance between O(0,0) and S(4,0) = 4. 
The height of the triangle corresponding to this base is the perpendicular distance from point P to the x-axis, which is simply the absolute value of the y-coordinate of P. 
Height = $|\alpha| = \left|\pm \frac{12}{5}\right| = \frac{12}{5}$. 
Area of $\Delta$POS = $\frac{1}{2} \times \text{base} \times \text{height}$ 
Area = $\frac{1}{2} \times 4 \times \frac{12}{5} = 2 \times \frac{12}{5} = \frac{24}{5}$. 
Step 4: Final Answer: 
The area of $\Delta$POS is $\frac{24}{5}$.