Ellipse Questions in English

(C) Given the focus $S(ae, 0) = (4, 0)$ and eccentricity $e = 4/5$.
Thus, $ae = 4 \implies a(4/5) = 4 \implies a = 5$.
Using the relation $b^2 = a^2(1 - e^2)$, we get $b^2 = 25(1 - 16/25) = 25(9/25) = 9$.
The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.
Since point $P(3, \alpha)$ lies on the ellipse, we substitute $x=3$ and $y=\alpha$ into the equation:
$\frac{3^2}{25} + \frac{\alpha^2}{9} = 1 \implies \frac{9}{25} + \frac{\alpha^2}{9} = 1 \implies \frac{\alpha^2}{9} = 1 - \frac{9}{25} = \frac{16}{25}$.
$\alpha^2 = \frac{16 \times 9}{25} \implies \alpha = \pm \frac{12}{5}$.
Taking $\alpha = 12/5$, the coordinates are $O(0, 0)$, $S(4, 0)$, and $P(3, 12/5)$.
The area of $\triangle POS = \frac{1}{2} |x_O(y_S - y_P) + x_S(y_P - y_O) + x_P(y_O - y_S)|$.
Area $= \frac{1}{2} |0(0 - 12/5) + 4(12/5 - 0) + 3(0 - 0)| = \frac{1}{2} |48/5| = 24/5$.

(B) For an ellipse with the major axis along the y-axis,the denominator of the $y^2$ term must be greater than the denominator of the $x^2$ term,and both must be positive: $f(3a+15) > f(a^2+7a+3) > 0$.
Since $f$ is a strictly decreasing function,$f(x_1) > f(x_2) \implies x_1 < x_2$.
Therefore,$3a+15 < a^2+7a+3$.
Rearranging the inequality gives $a^2 + 4a - 12 > 0$.
Factoring the quadratic expression,we get $(a+6)(a-2) > 0$.
This inequality holds when $a \in (-\infty, -6) \cup (2, \infty)$.
The set of values for $a$ is $R - [-6, 2]$.
Comparing this with $R - [\alpha, \beta]$,we get $\alpha = -6$ and $\beta = 2$.
Thus,$\alpha^2 + \beta^2 = (-6)^2 + (2)^2 = 36 + 4 = 40$.

(D) Given the ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a < b$.
Since $a < b$,the eccentricity $e$ is given by $e^2 = 1 - \frac{a^2}{b^2}$.
Given $e = \frac{\sqrt{5}}{3}$,so $e^2 = \frac{5}{9}$.
Thus,$1 - \frac{a^2}{b^2} = \frac{5}{9} \implies \frac{a^2}{b^2} = 1 - \frac{5}{9} = \frac{4}{9}$.
Let $a^2 = 4k$ and $b^2 = 9k$ for some constant $k > 0$.
The ellipse passes through $(4, 3)$,so $\frac{4^2}{a^2} + \frac{3^2}{b^2} = 1$.
Substituting the values,$\frac{16}{4k} + \frac{9}{9k} = 1 \implies \frac{4}{k} + \frac{1}{k} = 1 \implies \frac{5}{k} = 1 \implies k = 5$.
Therefore,$a^2 = 4(5) = 20$ and $b^2 = 9(5) = 45$.
This gives $a = \sqrt{20} = 2\sqrt{5}$ and $b = \sqrt{45} = 3\sqrt{5}$.
The length of the latus rectum for an ellipse with $a < b$ is $\frac{2a^2}{b}$.
Length $= \frac{2(20)}{3\sqrt{5}} = \frac{40}{3\sqrt{5}} = \frac{40\sqrt{5}}{3 \times 5} = \frac{8\sqrt{5}}{3}$.

(C) The parabola $P : y^2 = 4x$ has its focus at $(1, 0)$ and its latus rectum is the line $x = 1$.
Since the line segment joining the points of intersection of $P$ and $E$ is the latus rectum of both,the line $x = 1$ must be the latus rectum of the ellipse $E$.
For the ellipse,the latus rectum is at $x = ae$,so $ae = 1$.
The points of intersection are $(1, 2)$ and $(1, -2)$. Since these points lie on the ellipse,we substitute $x = 1$ and $y = 2$ into $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:
$\frac{1}{a^2} + \frac{4}{b^2} = 1$.
Using the relation $b^2 = a^2(1 - e^2)$,we substitute $b^2$:
$\frac{1}{a^2} + \frac{4}{a^2(1 - e^2)} = 1$.
Since $a = 1/e$,we have $a^2 = 1/e^2$,so $e^2 + \frac{4e^2}{1 - e^2} = 1$.
$e^2(1 - e^2) + 4e^2 = 1 - e^2 \implies e^2 - e^4 + 4e^2 = 1 - e^2 \implies e^4 - 6e^2 + 1 = 0$.
Solving for $e^2$ using the quadratic formula: $e^2 = \frac{6 \pm \sqrt{36 - 4}}{2} = 3 \pm 2\sqrt{2}$.
Since $e < 1$,$e^2 = 3 - 2\sqrt{2}$.
Then $e^2 + 2\sqrt{2} = (3 - 2\sqrt{2}) + 2\sqrt{2} = 3$.