On the ellipse x^28+y^24=1, let P be a point in the second quadrant such that the tangent …

Question

On the ellipse $\dfrac{x^2}8+\dfrac{y^2}4=1$, let $P$ be a point in the second quadrant such that the tangent at $P$ 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 $	riangle SPS'$, then the value of $(5-e^2)\cdot A$ is

Accepted Answer

$e^2=1-\dfrac48=\dfrac12$, $e=\dfrac1{\sqrt2}$; foci $S=(2,0),S'=(-2,0)$.

Tangent slope $=2$ (perpendicular to slope $-\frac12$): $y=2x+c$, $c^2=8(4)+4=36\Rightarrow c=6$ (for the second-quadrant point).

Point of tangency: $x_0=-\dfrac{a^2m}c=-\dfrac{16}6=-\dfrac83$, $y_0=\dfrac{b^2}c=\dfrac{4}6=\dfrac23$.

$$A=\frac12(SS')(y_0)=\frac12(4)\left(\frac23\right)=\frac43.$$

$$(5-e^2)A=\left(\frac92\right)\left(\frac43\right)=6.$$

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