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:
The problem requires us to first determine the specific equation of an ellipse from its given focus and eccentricity. Then, we must find the y-coordinate of a point on this ellipse given its x-coordinate. Finally, we calculate the area of the triangle formed by this point, the origin, and the given focus.
Step 2: Key Formula or Approach:
1. Ellipse Properties: For a standard ellipse with foci on the x-axis, the focus is at $(ae, 0)$, and the parameters are related by $b^2 = a^2(1-e^2)$.
2. Point on Ellipse: If a point $(x_0, y_0)$ lies on the ellipse, it must satisfy the ellipse's equation.
3. Area of a Triangle: With vertices at the origin, a point on the x-axis, and a third point, the area is easily calculated as $\frac{1}{2} \times \text{base} \times \text{height}$.
Step 3: Detailed Explanation:
Part 1: Find the Equation of the Ellipse
The focus is S(4, 0), so we have $ae = 4$.
The eccentricity is given as $e = \frac{4}{5}$.
Substituting $e$ into the focus equation:
$a \left(\frac{4}{5}\right) = 4 \implies a = 5$.
Now we find $b^2$ using the standard relation:
$b^2 = a^2(1-e^2) = 5^2 \left(1 - (\frac{4}{5})^2\right) = 25 \left(1 - \frac{16}{25}\right) = 25 \left(\frac{9}{25}\right) = 9$.
The equation of the ellipse is therefore:
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \] Part 2: Find the Coordinate of Point P
The point P(3, $\alpha$) lies on the ellipse, so we substitute its coordinates into the equation:
\[ \frac{3^2}{25} + \frac{\alpha^2}{9} = 1 \implies \frac{9}{25} + \frac{\alpha^2}{9} = 1 \] \[ \frac{\alpha^2}{9} = 1 - \frac{9}{25} = \frac{16}{25} \] \[ \alpha^2 = \frac{9 \times 16}{25} \implies \alpha = \pm \frac{12}{5} \] So, the point P is $(3, \pm 12/5)$.
Part 3: Calculate the Area of $\Delta$POS
The vertices of the triangle are P(3, $\pm 12/5$), O(0, 0), and S(4, 0).
We can take the segment OS, which lies on the x-axis, as the base of the triangle.
The length of the base is the distance from O to S, which is 4.
The height of the triangle is the perpendicular distance from P to the x-axis, which is the absolute value of the y-coordinate of P.
Height $= |\alpha| = |\pm \frac{12}{5}| = \frac{12}{5}$.
The area of the triangle is:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times \frac{12}{5} = \frac{24}{5} \] Step 4: Final Answer:
The area of $\Delta$POS is $\frac{24}{5}$.