On the ellipse \(\frac{x^2}{8} + \frac{y^2}{4} = 1\) let \(P\) be a point in the second quadrant such that the tangent at \(P\) to the ellipse is perpendicular to the line \(x + 2y = 0\). Let \(S\) and \(S'\) be the foci of the ellipse and \(e\) be its eccentricity. If \(A\) is the area of the triangle \(SPS'\) then, the value of \((5 - e^2) \cdot A\) is :
Show Hint
The area of a triangle with base on the x-axis (like \(SS'\)) is simply \(ae \cdot y_P\). This shortcut avoids using the general determinant formula for area.
Step 1: Understanding the Concept:
We determine the coordinates of point \(P\) using the slope of the tangent. Then, calculate the eccentricity and the locations of the foci. Finally, compute the area of the triangle and the required expression value. Step 2: Key Formula or Approach:
1. Tangent slope \(m = -1/\text{slope of line}\).
2. Tangent to \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) at \((x_1, y_1)\) is \(\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1\).
3. \(e^2 = 1 - b^2/a^2\).
4. Area of \(\triangle SPS' = \frac{1}{2} \cdot |x_S - x_{S'}| \cdot |y_P|\). Step 3: Detailed Explanation:
Given line \(x + 2y = 0\) has slope \(-1/2\). The tangent is perpendicular, so its slope is \(m = 2\).
Ellipse: \(a^2 = 8, b^2 = 4\). \(e^2 = 1 - 4/8 = 1/2\).
The tangent is \(y = mx \pm \sqrt{a^2m^2 + b^2} = 2x \pm \sqrt{8(4) + 4} = 2x \pm 6\).
For the second quadrant (\(x<0, y>0\)), take \(y = 2x + 6\).
Comparing with \(\frac{xx_1}{8} + \frac{yy_1}{4} = 1 \implies x\left(-\frac{x_1}{2y_1}\right) + \frac{4}{y_1} = y\).
Slope \(m = -x_1/2y_1 = 2 \implies x_1 = -4y_1\).
Substitute into ellipse: \(\frac{16y_1^2}{8} + \frac{y_1^2}{4} = 1 \implies 2y_1^2 + 0.25y_1^2 = 1 \implies 9y_1^2 = 4 \implies y_1 = 2/3\).
So, height of triangle \(h = 2/3\).
Distance between foci \(SS' = 2ae = 2\sqrt{8}\frac{1}{\sqrt{2}} = 2(2) = 4\).
Area \(A = \frac{1}{2} \cdot 4 \cdot \frac{2}{3} = \frac{4}{3}\).
Value of \((5 - e^2)A = (5 - 0.5) \cdot \frac{4}{3} = \frac{9}{2} \cdot \frac{4}{3} = 6\). Step 4: Final Answer:
The required value is 6.
