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 $e^2 + 2\sqrt{2} = (3 - 2\sqrt{2}) + 2\sqrt{2} = 3$.

(A) The equation of the ellipse is $9x^2 + 16y^2 = 288$. Dividing by $288$, we get $\frac{x^2}{32} + \frac{y^2}{18} = 1$. Here $a^2 = 32$ and $b^2 = 18$.
The equation of a tangent with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.
Since the tangent makes equal intercepts on the axes, its slope must be $m = -1$ or $m = 1$. Given it makes intercepts on the axes, we consider the form $\frac{x}{c} + \frac{y}{c} = 1$, which implies $y = -x + c$, so $m = -1$.
The condition for tangency is $c^2 = a^2m^2 + b^2$. Substituting $m = -1$, $c^2 = 32(-1)^2 + 18 = 32 + 18 = 50$.
Thus, $c = \pm \sqrt{50} = \pm 5\sqrt{2}$.
The intercepts are $A(5\sqrt{2}, 0)$ and $B(0, 5\sqrt{2})$.
The area of $\triangle OAB = \frac{1}{2} \times |5\sqrt{2}| \times |5\sqrt{2}| = \frac{1}{2} \times 50 = 25$ sq. units.

(A) For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the equation of a tangent with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.
Here, $a^2 = 16$, $b^2 = 9$, and the inclination is $30^\circ$, so $m = \tan(30^\circ) = \frac{1}{\sqrt{3}}$.
Substituting these values into the formula:
$y = \frac{1}{\sqrt{3}}x \pm \sqrt{16(\frac{1}{\sqrt{3}})^2 + 9}$
$y = \frac{x}{\sqrt{3}} \pm \sqrt{\frac{16}{3} + 9}$
$y = \frac{x}{\sqrt{3}} \pm \sqrt{\frac{16 + 27}{3}}$
$y = \frac{x}{\sqrt{3}} \pm \sqrt{\frac{43}{3}}$
Multiply by $\sqrt{3}$:
$\sqrt{3}y = x \pm \sqrt{43}$
$x - \sqrt{3}y \pm \sqrt{43} = 0$.

(D) The given equation of the ellipse is $9x^2 + 16y^2 = 144$. Dividing by $144$, we get $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we have $a^2 = 16$ and $b^2 = 9$.
The condition for the line $y = mx + c$ to touch the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$.
Rewrite the line $3x + 4y + k = 0$ as $4y = -3x - k$, which gives $y = -\frac{3}{4}x - \frac{k}{4}$.
Here, $m = -\frac{3}{4}$ and $c = -\frac{k}{4}$.
Substituting these values into the condition: $(-\frac{k}{4})^2 = 16(-\frac{3}{4})^2 + 9$.
$\frac{k^2}{16} = 16(\frac{9}{16}) + 9$.
$\frac{k^2}{16} = 9 + 9 = 18$.
$k^2 = 18 \times 16 = 288$.
$k = \pm \sqrt{288} = \pm 12\sqrt{2}$.

(A) The equation of the ellipse is $3x^2 + 4y^2 = 12$, which can be written as $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Here $a^2 = 4$ and $b^2 = 3$.
The equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.
Given $m = -1/2$, the equation becomes $y = -\frac{1}{2}x \pm \sqrt{4(-\frac{1}{2})^2 + 3} = -\frac{1}{2}x \pm \sqrt{1 + 3} = -\frac{1}{2}x \pm 2$.
For the $X$-intercept $(A)$, set $y = 0$: $0 = -\frac{1}{2}x \pm 2 \implies x = \pm 4$. So $A = (\pm 4, 0)$.
For the $Y$-intercept $(B)$, set $x = 0$: $y = \pm 2$. So $B = (0, \pm 2)$.
The area of $\Delta AOB = \frac{1}{2} \times |x_A| \times |y_B| = \frac{1}{2} \times 4 \times 2 = 4 \text{ sq. units}$.

(A) The equation of the ellipse is $4x^2 + 9y^2 = 1$, which can be written as $\frac{x^2}{(1/2)^2} + \frac{y^2}{(1/3)^2} = 1$.
Here, $a^2 = 1/4$ and $b^2 = 1/9$. Since $a^2 > b^2$, the major axis is along the $X$-axis.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1/9}{1/4}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.
The foci $S$ and $S'$ are at $(\pm ae, 0)$, so $S = (\frac{1}{2} \cdot \frac{\sqrt{5}}{3}, 0) = (\frac{\sqrt{5}}{6}, 0)$ and $S' = (-\frac{\sqrt{5}}{6}, 0)$.
The ellipse intersects the positive $Y$-axis at $A(0, b) = (0, 1/3)$.
The base of $\Delta SAS'$ is the distance $SS' = \frac{\sqrt{5}}{6} - (-\frac{\sqrt{5}}{6}) = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$.
The height of the triangle is the $y$-coordinate of $A$, which is $1/3$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{\sqrt{5}}{3} \times \frac{1}{3} = \frac{\sqrt{5}}{18}$ sq. units.

(D) The foci of the ellipse are $(\pm ae, 0)$. Since the line $3x + 2y - 6 = 0$ is a focal chord, it must pass through one of the foci $(\pm ae, 0)$.
Given $e = 2/3$, the foci are $(\pm \frac{2a}{3}, 0)$.
Case $1$: Line passes through $(\frac{2a}{3}, 0) \implies 3(\frac{2a}{3}) + 2(0) - 6 = 0 \implies 2a = 6 \implies a = 3$.
Case $2$: Line passes through $(-\frac{2a}{3}, 0) \implies 3(-\frac{2a}{3}) + 2(0) - 6 = 0 \implies -2a = 6 \implies a = -3$ (not possible as $a > 0$).
Using $b^2 = a^2(1 - e^2)$, we get $b^2 = 3^2(1 - (2/3)^2) = 9(1 - 4/9) = 9(5/9) = 5$.
Thus, $a^2 + b^2 = 3^2 + 5 = 9 + 5 = 14$.

(A) Step $1$: Rewrite the equation by completing the square for $x$ and $y$ terms.
$7(x^2 - 2x) + 16(y^2 + 4y) = 377$
Step $2$: Add constants to complete the squares.
$7(x^2 - 2x + 1) + 16(y^2 + 4y + 4) = 377 + 7(1) + 16(4)$
$7(x - 1)^2 + 16(y + 2)^2 = 377 + 7 + 64 = 448$
Step $3$: Divide by $448$ to get the standard form.
$\frac{7(x - 1)^2}{448} + \frac{16(y + 2)^2}{448} = 1$
$\frac{(x - 1)^2}{64} + \frac{(y + 2)^2}{28} = 1$
Step $4$: Identify $a^2$ and $b^2$. Here $a^2 = 64$ and $b^2 = 28$.
Step $5$: Calculate eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}}$.
$e = \sqrt{1 - \frac{28}{64}} = \sqrt{\frac{64 - 28}{64}} = \sqrt{\frac{36}{64}} = \frac{6}{8} = \frac{3}{4}$.

(A) The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{9} = 1$, where $a^2 = 25$ and $b^2 = 9$. Thus, $a = 5$ and $b = 3$.
Any point on the ellipse can be represented as $(a \cos \theta, b \sin \theta) = (5 \cos \theta, 3 \sin \theta)$.
The distance of this point from the center $(0, 0)$ is given as $5$.
Using the distance formula: $\sqrt{(5 \cos \theta - 0)^2 + (3 \sin \theta - 0)^2} = 5$.
Squaring both sides: $25 \cos^2 \theta + 9 \sin^2 \theta = 25$.
Substitute $\cos^2 \theta = 1 - \sin^2 \theta$: $25(1 - \sin^2 \theta) + 9 \sin^2 \theta = 25$.
$25 - 25 \sin^2 \theta + 9 \sin^2 \theta = 25$.
$-16 \sin^2 \theta = 0$, which implies $\sin \theta = 0$.
Therefore, $\theta = 0$ or $\pi$.

(C) The equation of the ellipse is $16x^2 + 25y^2 = 400$. Dividing by $400$, we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$.
Here, $a^2 = 25$ and $b^2 = 16$, so $a = 5$ and $b = 4$.
The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
The foci are $S(ae, 0) = (3, 0)$ and $S'(-ae, 0) = (-3, 0)$.
Let $P = (x_0, y_0)$. The area of $\Delta PSS'$ is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (SS') \times |y_0| = 9$.
The distance $SS' = 2ae = 2(3) = 6$.
So, $\frac{1}{2} \times 6 \times |y_0| = 9 \implies 3|y_0| = 9 \implies |y_0| = 3$.
Since $P$ lies on the ellipse, $16x_0^2 + 25(3)^2 = 400$.
$16x_0^2 + 225 = 400 \implies 16x_0^2 = 175 \implies x_0^2 = \frac{175}{16}$.
$x_0 = \pm \frac{\sqrt{25 \times 7}}{4} = \pm \frac{5\sqrt{7}}{4}$.
Thus, the abscissa is $\frac{5\sqrt{7}}{4}$.