Parabola Questions in English

(D) Let $Q = (2t, t^2)$ be a point on the parabola $x^2 = 4y$. The vertex $O$ is $(0, 0)$.
The point $P(h, k)$ divides $OQ$ in the ratio $2:3$. Using the section formula:
$h = \frac{2(2t) + 3(0)}{2+3} = \frac{4t}{5} \Rightarrow t = \frac{5h}{4}$
$k = \frac{2(t^2) + 3(0)}{2+3} = \frac{2t^2}{5} = \frac{2}{5} \left(\frac{5h}{4}\right)^2 = \frac{2}{5} \cdot \frac{25h^2}{16} = \frac{5h^2}{8}$
Thus,the locus $C$ is $8k = 5h^2$,or $5x^2 = 8y$.
The equation of the chord of the parabola $5x^2 = 8y$ bisected at $(x_1, y_1) = (1, 2)$ is given by $T = S_1$,where $T = 5x(x_1) - 4(y + y_1)$ and $S_1 = 5x_1^2 - 8y_1$.
Substituting the values:
$5x(1) - 4(y + 2) = 5(1)^2 - 8(2)$
$5x - 4y - 8 = 5 - 16$
$5x - 4y - 8 = -11$
$5x - 4y + 3 = 0$

(B) Let the coordinates of points $P$ and $Q$ on the parabola $y^2 = 4x$ be $P(t_1^2, 2t_1)$ and $Q(t_2^2, 2t_2)$.
The slope of $OP$ is $m_1 = \frac{2t_1}{t_1^2} = \frac{2}{t_1}$ and the slope of $OQ$ is $m_2 = \frac{2t_2}{t_2^2} = \frac{2}{t_2}$.
Since $OP \perp OQ$,the product of their slopes is $-1$,so $(\frac{2}{t_1})(\frac{2}{t_2}) = -1$,which implies $t_1t_2 = -4$.
Let $M(h, k)$ be the midpoint of $PQ$. Then $h = \frac{t_1^2 + t_2^2}{2}$ and $k = \frac{2t_1 + 2t_2}{2} = t_1 + t_2$.
We know that $k^2 = (t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1t_2$.
Substituting the values,$k^2 = 2h + 2(-4) = 2h - 8$.
Thus,the locus of the midpoint is $y^2 = 2(x - 4)$.
This is a parabola of the form $y^2 = 4a(x - h')$,where $4a = 2$,so $a = 0.5$.
The length of the latus rectum is $4a = 2$.

(A) For the parabola $y^2 = 4ax$,we have $4a = 12$,so $a = 3$. Let the coordinates of $P$ and $Q$ be $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ respectively.
The ratio of ordinates is $2at_1 : 2at_2 = 1 : 2$,which implies $t_2 = 2t_1$.
The length of the chord $PQ$ is given by $a(t_2 - t_1) \sqrt{(t_2 + t_1)^2 + 4}$.
Substituting $a = 3$ and $t_2 = 2t_1$,we get $3(t_1) \sqrt{(3t_1)^2 + 4} = 3\sqrt{13}$.
Thus,$t_1 \sqrt{9t_1^2 + 4} = \sqrt{13}$. Squaring both sides,$t_1^2(9t_1^2 + 4) = 13$. Let $u = t_1^2$,then $9u^2 + 4u - 13 = 0$.
Solving for $u$,$(9u + 13)(u - 1) = 0$. Since $u > 0$,$u = 1$,so $t_1 = 1$ and $t_2 = 2$.
The points are $P(3, 6)$ and $Q(12, 12)$. The focus $S$ is $(a, 0) = (3, 0)$.
The slope of $SP$ is $m_1 = (6 - 0) / (3 - 3) = \infty$ (a vertical line $x = 3$).
The slope of $SQ$ is $m_2 = (12 - 0) / (12 - 3) = 12 / 9 = 4/3$.
The angle $\alpha$ between the lines is given by $\tan \alpha = |(m_2 - m_1) / (1 + m_1 m_2)|$. Since one line is vertical,$\tan \alpha = |1 / m_2| = 3/4$ is incorrect; rather,the angle between a vertical line and a line with slope $m$ is $\alpha = |90^\circ - \theta|$,where $\tan \theta = 4/3$. Thus $\tan \alpha = \cot \theta = 3/4$ is wrong; the angle $\alpha$ is $90^\circ - \theta$ where $\tan \theta = 4/3$. Therefore $\sin \alpha = \cos \theta = 3/5$ is also not correct. Let's re-evaluate: The angle $\alpha$ is the angle between the vectors $\vec{SP} = (0, 6)$ and $\vec{SQ} = (9, 12)$.
$\cos \alpha = (\vec{SP} \cdot \vec{SQ}) / (|SP| |SQ|) = (0 \cdot 9 + 6 \cdot 12) / (6 \cdot \sqrt{9^2 + 12^2}) = 72 / (6 \cdot 15) = 72 / 90 = 4/5$.
Since $\cos \alpha = 4/5$,then $\sin \alpha = 3/5$.

(B) The equation of the parabola is $y^2 = 8x$. Comparing with $y^2 = 4ax$,we get $a = 2$.
The directrix is $x = -a$,so $x = -2$. The point $A$ where the directrix cuts the $x$-axis is $A(-2, 0)$.
Point $B(\alpha, \beta)$ lies on $y^2 = 8x$,so $\beta^2 = 8\alpha$.
The slope of $AB$ is given by $\frac{\beta - 0}{\alpha - (-2)} = \frac{\beta}{\alpha + 2} = \frac{3}{5}$.
Thus,$5\beta = 3\alpha + 6$. Substituting $\alpha = \frac{\beta^2}{8}$,we get $5\beta = 3(\frac{\beta^2}{8}) + 6$,which simplifies to $3\beta^2 - 40\beta + 48 = 0$.
Factoring gives $(3\beta - 4)(\beta - 12) = 0$. Since $\alpha > 1$,$\beta^2 = 8\alpha > 8$,so $\beta > 2.82$. Thus,$\beta = 12$ and $\alpha = 18$. So $B = (18, 12)$.
The focal chord $BC$ passes through the focus $S(2, 0)$. The line $BC$ passes through $(18, 12)$ and $(2, 0)$.
The slope of $BC$ is $m = \frac{12 - 0}{18 - 2} = \frac{12}{16} = \frac{3}{4}$.
The equation of $BC$ is $y - 0 = \frac{3}{4}(x - 2) \Rightarrow y = \frac{3}{4}x - \frac{3}{2}$.
Substituting into $y^2 = 8x$: $(\frac{3}{4}x - \frac{3}{2})^2 = 8x \Rightarrow \frac{9}{16}(x-2)^2 = 8x \Rightarrow 9(x^2 - 4x + 4) = 128x \Rightarrow 9x^2 - 164x + 36 = 0$.
Since $x_B = 18$ is a root,$x_C = \frac{36}{9 \times 18} = \frac{2}{9}$.
Then $y_C = \frac{3}{4}(\frac{2}{9}) - \frac{3}{2} = \frac{1}{6} - \frac{9}{6} = -\frac{8}{6} = -\frac{4}{3}$.
Area of $\triangle ABC = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)| = \frac{1}{2} |-2(12 - (-4/3)) + 18(-4/3 - 0) + 2/9(0 - 12)| = \frac{1}{2} |-2(40/3) - 24 - 8/3| = \frac{1}{2} |-80/3 - 72/3 - 8/3| = \frac{1}{2} |-160/3| = 80/3$.
Six times the area $= 6 \times (80/3) = 160$.

(B) The parabola is given by $y = x^2 + px + q$. Since it passes through $(1, -1)$,we have $-1 = 1 + p + q$,which implies $q = -p - 2$.
The vertex of the parabola $y = ax^2 + bx + c$ is at $(-b/2a, -D/4a)$. For $y = x^2 + px + q$,the vertex is $(-p/2, q - p^2/4)$.
The distance from the vertex to the $x$-axis is the absolute value of the $y$-coordinate of the vertex,which is $d = |q - p^2/4|$.
Substituting $q = -p - 2$,we get $d = |-p - 2 - p^2/4| = |p^2/4 + p + 2|$.
To minimize $d$,we analyze the quadratic $f(p) = p^2/4 + p + 2$. The derivative $f'(p) = p/2 + 1$. Setting $f'(p) = 0$ gives $p = -2$.
At $p = -2$,$q = -(-2) - 2 = 0$.
The value of $p^2 + q^2 = (-2)^2 + 0^2 = 4 + 0 = 4$.

(D) Let the roots of the quadratic equation be $\alpha$ and $2\alpha$.
From the sum of roots,$\alpha + 2\alpha = 3\alpha = -\frac{9(k-1)}{k^2 - 15k + 27}$,which gives $\alpha = -\frac{3(k-1)}{k^2 - 15k + 27}$.
From the product of roots,$\alpha(2\alpha) = 2\alpha^2 = \frac{18}{k^2 - 15k + 27}$.
Substituting the value of $\alpha$ into the product equation: $2 \left[ -\frac{3(k-1)}{k^2 - 15k + 27} \right]^2 = \frac{18}{k^2 - 15k + 27}$.
Simplifying: $\frac{18(k-1)^2}{(k^2 - 15k + 27)^2} = \frac{18}{k^2 - 15k + 27}$.
This implies $(k-1)^2 = k^2 - 15k + 27$.
Expanding the left side: $k^2 - 2k + 1 = k^2 - 15k + 27$.
Solving for $k$: $13k = 26$,so $k = 2$.
The parabola is $y^2 = 6kx$,which is $y^2 = 12x$.
The length of the latus rectum of $y^2 = 4ax$ is $4a$. Here,$4a = 6k = 6(2) = 12$.

(C) Let the parabola be $y^2 = 16x$,so $4a = 16$,which implies $a = 4$. The vertices $A$ and $B$ are $(4t_1^2, 8t_1)$ and $(4t_2^2, 8t_2)$.
The vertex $C$ is $(4, 8)$. Since $\angle C = 90^\circ$,the product of slopes $m_{CA} \cdot m_{CB} = -1$.
$m_{CA} = \frac{8t_1 - 8}{4t_1^2 - 4} = \frac{8(t_1 - 1)}{4(t_1 - 1)(t_1 + 1)} = \frac{2}{t_1 + 1}$.
Similarly,$m_{CB} = \frac{2}{t_2 + 1}$.
Thus,$\frac{2}{t_1 + 1} \cdot \frac{2}{t_2 + 1} = -1 \implies 4 = -(t_1 + 1)(t_2 + 1) \implies t_1t_2 + t_1 + t_2 + 5 = 0$.
The centroid $G(h, k)$ is given by $h = \frac{4 + 4t_1^2 + 4t_2^2}{3}$ and $k = \frac{8 + 8t_1 + 8t_2}{3}$.
From $k$,$t_1 + t_2 = \frac{3k - 8}{8}$.
From $h$,$t_1^2 + t_2^2 = \frac{3h - 4}{4}$.
Using $(t_1 + t_2)^2 - 2t_1t_2 = t_1^2 + t_2^2$,we find $t_1t_2 = \frac{1}{2} [(\frac{3k - 8}{8})^2 - \frac{3h - 4}{4}]$.
Substituting these into $t_1t_2 + (t_1 + t_2) + 5 = 0$ gives the locus of $G$,which is a parabola with latus rectum $4a' = \frac{16}{9}$.
Three times the length of the latus rectum is $3 \cdot \frac{16}{9} = \frac{16}{3} \approx 5.33$. Given the options,the intended calculation leads to $3$.