Parabola Questions in English

(A) Let $f(x) = x^2 - 3x + 3$.
Completing the square,we get $f(x) = (x - 3/2)^2 + 3/4$.
The vertex of the parabola is at $x = 3/2$.
Since the interval is $(-3, 3/2)$,the function $f(x)$ is strictly decreasing on this interval.
As $x$ approaches $3/2$ from the left,$f(x)$ approaches $3/4$.
Since $3/2$ is not included in the interval,the function does not attain the value $3/4$,but it is the infimum (greatest lower bound) of the function on this interval.

(A) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Substituting the given points $(am_1^2, 2am_1)$ and $(am_2^2, 2am_2)$:
$d = \sqrt{(am_2^2 - am_1^2)^2 + (2am_2 - 2am_1)^2}$
$d = \sqrt{a^2(m_2^2 - m_1^2)^2 + 4a^2(m_2 - m_1)^2}$
$d = \sqrt{a^2(m_2 - m_1)^2(m_2 + m_1)^2 + 4a^2(m_2 - m_1)^2}$
$d = \sqrt{a^2(m_2 - m_1)^2 [(m_1 + m_2)^2 + 4]}$
$d = a|m_1 - m_2|\sqrt{(m_1 + m_2)^2 + 4}$.

(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the given vertices:
$\Delta = \frac{1}{2} |am_1^2(2am_2 - 2am_3) + am_2^2(2am_3 - 2am_1) + am_3^2(2am_1 - 2am_2)|$
$\Delta = \frac{1}{2} \cdot 2a^2 |m_1^2(m_2 - m_3) + m_2^2(m_3 - m_1) + m_3^2(m_1 - m_2)|$
Using the identity $m_1^2(m_2 - m_3) + m_2^2(m_3 - m_1) + m_3^2(m_1 - m_2) = -(m_1 - m_2)(m_2 - m_3)(m_3 - m_1)$,
$\Delta = a^2 |-(m_1 - m_2)(m_2 - m_3)(m_3 - m_1)| = a^2 |(m_1 - m_2)(m_2 - m_3)(m_3 - m_1)|$.

(A) Let the point be $P(x, y)$.
The distance of $P(x, y)$ from the point $(a, 0)$ is $\sqrt{(x - a)^2 + (y - 0)^2}$.
The distance of $P(x, y)$ from the line $x + a = 0$ is $|x + a|$.
According to the problem,these distances are equal:
$\sqrt{(x - a)^2 + y^2} = |x + a|$
Squaring both sides,we get:
$(x - a)^2 + y^2 = (x + a)^2$
$x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2$
$y^2 = 4ax$
This is the standard equation of a parabola.

(A) Let the point be $P(x, y)$.
The distance from $P(x, y)$ to the point $(4, 2)$ is $\sqrt{(x - 4)^2 + (y - 2)^2}$.
The distance from $P(x, y)$ to the $x$-axis is $|y|$.
Since the point is equidistant,we have $\sqrt{(x - 4)^2 + (y - 2)^2} = |y|$.
Squaring both sides,we get $(x - 4)^2 + (y - 2)^2 = y^2$.
Expanding the terms: $x^2 - 8x + 16 + y^2 - 4y + 4 = y^2$.
Simplifying,we get $x^2 - 8x - 4y + 20 = 0$.

(B) Let $x = (u \cos \alpha)t$ and $y = (u \sin \alpha)t - \frac{1}{2}gt^2$.
From the first equation,we have $t = \frac{x}{u \cos \alpha}$.
Substituting this value of $t$ into the equation for $y$:
$y = (u \sin \alpha) \left( \frac{x}{u \cos \alpha} \right) - \frac{1}{2}g \left( \frac{x}{u \cos \alpha} \right)^2$
$y = x \tan \alpha - \frac{g x^2}{2 u^2 \cos^2 \alpha}$.
This equation is of the form $y = ax^2 + bx + c$,which represents a parabola.

(C) Let $P = (1, 0)$ and $Q = (h, k)$ be a point on the parabola $y^2 = 8x$,so $k^2 = 8h$.
Let $(x, y)$ be the midpoint of $PQ$.
Then $x = \frac{h + 1}{2}$ and $y = \frac{k + 0}{2}$.
This implies $h = 2x - 1$ and $k = 2y$.
Substituting these into the equation $k^2 = 8h$:
$(2y)^2 = 8(2x - 1)$
$4y^2 = 16x - 8$
Dividing by $4$,we get $y^2 = 4x - 2$,which is $y^2 - 4x + 2 = 0$.

(C) Given points are $A(at^2, 2at)$,$B(a/t^2, -2a/t)$,and $C(a, 0)$.
$CA = \sqrt{(at^2 - a)^2 + (2at - 0)^2} = \sqrt{a^2(t^2 - 1)^2 + 4a^2t^2} = a\sqrt{t^4 - 2t^2 + 1 + 4t^2} = a\sqrt{(t^2 + 1)^2} = a(t^2 + 1)$.
$CB = \sqrt{(\frac{a}{t^2} - a)^2 + (-\frac{2a}{t} - 0)^2} = \sqrt{a^2(\frac{1 - t^2}{t^2})^2 + \frac{4a^2}{t^2}} = a\sqrt{\frac{1 - 2t^2 + t^4 + 4t^2}{t^4}} = a\sqrt{\frac{(1 + t^2)^2}{t^4}} = a\frac{1 + t^2}{t^2}$.
$H.M.$ of $CA$ and $CB$ is given by $\frac{2(CA)(CB)}{CA + CB}$.
$H.M. = \frac{2 \cdot a(t^2 + 1) \cdot a(\frac{1 + t^2}{t^2})}{a(t^2 + 1) + a(\frac{1 + t^2}{t^2})} = \frac{2a^2(t^2 + 1)^2 / t^2}{a(t^2 + 1)(1 + 1/t^2)} = \frac{2a(t^2 + 1)^2 / t^2}{(t^2 + 1)(\frac{t^2 + 1}{t^2})} = \frac{2a(t^2 + 1)^2 / t^2}{(t^2 + 1)^2 / t^2} = 2a$.
Thus,$2a$ is the $H.M.$ of $CA$ and $CB$.

(B) Three points $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are collinear if the slope of the line segment connecting the first two points is equal to the slope of the line segment connecting the second and third points.
Slope $m_1$ between $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ is $\frac{2at_2 - 2at_1}{at_2^2 - at_1^2} = \frac{2a(t_2 - t_1)}{a(t_2 - t_1)(t_2 + t_1)} = \frac{2}{t_1 + t_2}$.
Slope $m_2$ between $(at_2^2, 2at_2)$ and $(a, 0)$ is $\frac{0 - 2at_2}{a - at_2^2} = \frac{-2at_2}{a(1 - t_2^2)} = \frac{-2t_2}{1 - t_2^2}$.
Equating the slopes: $\frac{2}{t_1 + t_2} = \frac{-2t_2}{1 - t_2^2}$.
$1 - t_2^2 = -t_2(t_1 + t_2)$.
$1 - t_2^2 = -t_1t_2 - t_2^2$.
$1 = -t_1t_2$,which implies $t_1t_2 = -1$.

(C) The equation of the parabola is $y^2 = 4ax$. The vertex is at $O(0, 0)$.
Let the double ordinate be at $x = h$. Then $y^2 = 4ah$,so $y = \pm 2\sqrt{ah}$.
The length of the double ordinate is $2 \times 2\sqrt{ah} = 4\sqrt{ah}$.
Given the length is $8a$,we have $4\sqrt{ah} = 8a$,which implies $\sqrt{ah} = 2a$,so $ah = 4a^2$,hence $h = 4a$.
The ends of the double ordinate are $P(4a, 4a)$ and $Q(4a, -4a)$.
The slope of $OP$ is $m_1 = \frac{4a - 0}{4a - 0} = 1$.
The slope of $OQ$ is $m_2 = \frac{-4a - 0}{4a - 0} = -1$.
Since $m_1 \times m_2 = 1 \times (-1) = -1$,the lines $OP$ and $OQ$ are perpendicular to each other.
Therefore,the angle between them is $90^\circ$.

(A) Let the coordinates of $P$ be $(h, k)$ and $Q$ be $(h, -k)$. Since $P$ lies on the parabola $y^2 = 4ax$,we have $k^2 = 4ah$.
Let $(x, y)$ be a point of trisection of $PQ$. The points of trisection divide $PQ$ in the ratio $1:2$ or $2:1$.
For the point dividing $PQ$ in ratio $1:2$,the $y$-coordinate is $y = \frac{1(-k) + 2(k)}{1+2} = \frac{k}{3}$,so $k = 3y$.
The $x$-coordinate is $x = h$.
Substituting $k = 3y$ and $h = x$ into the parabola equation $k^2 = 4ah$,we get $(3y)^2 = 4ax$,which simplifies to $9y^2 = 4ax$.

(C) The vertex of the parabola is at the origin $(0, 0)$.
Given the directrix is $x + 5 = 0$,which implies $x = -5$.
For a parabola with vertex at the origin and directrix $x = -a$,the value of $a$ is $5$.
The standard equation of this parabola is $y^2 = 4ax$.
Substituting $a = 5$,we get $y^2 = 4(5)x = 20x$.
The length of the latus rectum is given by $4a$.
Therefore,the length of the latus rectum is $4 \times 5 = 20$.

(B) The distance $d$ between the focus $(x_1, y_1)$ and the directrix $ax + by + c = 0$ is given by $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$.
Here,the focus is $(3, -4)$ and the directrix is $x + y - 2 = 0$.
So,$d = \frac{|1(3) + 1(-4) - 2|}{\sqrt{1^2 + 1^2}} = \frac{|3 - 4 - 2|}{\sqrt{2}} = \frac{|-3|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.
The length of the latus rectum of a parabola is $4$ times the distance from the focus to the directrix,but in standard geometry,the distance from the focus to the vertex is $a$,and the distance from the focus to the directrix is $2a$. The length of the latus rectum is $4a$.
Therefore,the length of the latus rectum $= 2 \times (\text{distance between focus and directrix}) = 2 \times \frac{3}{\sqrt{2}} = 3\sqrt{2}$.

(D) The equation of the parabola is $y^2 = 6x$.
Given the abscissa (x-coordinate) is $24$,we substitute $x = 24$ into the equation:
$y^2 = 6(24) = 144$
$y = \pm 12$.
Thus,the points on the parabola are $(24, 12)$ and $(24, -12)$.
The vertex of the parabola $y^2 = 6x$ is $(0, 0)$.
The equation of the line joining $(0, 0)$ and $(24, 12)$ is $y - 0 = \frac{12 - 0}{24 - 0}(x - 0)$ $\Rightarrow y = \frac{1}{2}x$ $\Rightarrow 2y = x$.
The equation of the line joining $(0, 0)$ and $(24, -12)$ is $y - 0 = \frac{-12 - 0}{24 - 0}(x - 0)$ $\Rightarrow y = -\frac{1}{2}x$ $\Rightarrow 2y = -x$.
Combining these,we get $2y = \pm x$,which can also be written as $x \pm 2y = 0$.

(A) Let the point on the parabola be $(x_1, y_1)$.
Given that the ordinate is three times the abscissa,we have $y_1 = 3x_1$.
Since the point lies on the parabola $y^2 = 36x$,we substitute $y_1$ into the equation:
$(3x_1)^2 = 36x_1$
$9x_1^2 = 36x_1$
$9x_1^2 - 36x_1 = 0$
$9x_1(x_1 - 4) = 0$
This gives $x_1 = 0$ or $x_1 = 4$.
If $x_1 = 0$,then $y_1 = 3(0) = 0$.
If $x_1 = 4$,then $y_1 = 3(4) = 12$.
Thus,the points are $(0, 0)$ and $(4, 12)$.

(B) For a parabola $y^2 = 4ax$,the focal distance of a point $(x, y)$ is given by $x + a$.
Given $y^2 = 12x$,we have $4a = 12$,so $a = 3$.
The focal distance is $x + a = 4$.
Substituting $a = 3$,we get $x + 3 = 4$,which implies $x = 1$.
Substituting $x = 1$ into the equation $y^2 = 12x$,we get $y^2 = 12(1) = 12$.
Thus,$y = \pm \sqrt{12} = \pm 2\sqrt{3}$.
Therefore,the points are $(1, 2\sqrt{3})$ and $(1, -2\sqrt{3})$.

(B) The given equation of the parabola is $5y^2 = 4x$,which can be rewritten as $y^2 = \frac{4}{5}x$.
Comparing this with the standard form $y^2 = 4ax$,we get $4a = \frac{4}{5}$,so $a = \frac{1}{5}$.
The focus of the parabola is $(a, 0) = (\frac{1}{5}, 0)$.
The latus rectum is a line perpendicular to the axis of the parabola passing through the focus,so its equation is $x = a = \frac{1}{5}$.
Substituting $x = \frac{1}{5}$ into the parabola equation $y^2 = \frac{4}{5}x$,we get $y^2 = \frac{4}{5} \times \frac{1}{5} = \frac{4}{25}$.
Thus,$y = \pm \frac{2}{5}$.
Therefore,the coordinates of the extremities of the latus rectum are $(\frac{1}{5}, \frac{2}{5})$ and $(\frac{1}{5}, -\frac{2}{5})$.

(B) Since the vertex is at the origin and the $y$-axis is the axis of the parabola,the equation is of the form $x^2 = 4ay$.
Given that the parabola passes through the point $(-4, -2)$,we substitute these coordinates into the equation:
$(-4)^2 = 4a(-2)$
$16 = -8a$
$a = -2$.
Thus,the equation of the parabola is $x^2 = -8y$.
The length of the latus rectum is given by $|4a|$.
$|4a| = |4(-2)| = |-8| = 8$.

(D) The given equation of the parabola is $x^2 = -16y$.
Comparing this with the standard form $x^2 = -4ay$,we get $4a = 16$,which implies $a = 4$.
Since the parabola opens downwards,the focus is given by the coordinates $(0, -a)$.
Substituting $a = 4$,the focus is $(0, -4)$.

(C) The equation of the parabola is $y^2 = 4ax$.
Since the parabola passes through the point $(-3, 2)$,we substitute $x = -3$ and $y = 2$ into the equation:
$(2)^2 = 4a(-3)$
$4 = -12a$
$a = -4/12 = -1/3$.
The length of the latus rectum is given by $|4a|$.
$|4a| = |4 \times (-1/3)| = |-4/3| = 4/3$.
Thus,the length of the latus rectum is $4/3$.

(C) The given equation of the parabola is $x^2 = -8y$.
Comparing this with the standard form $x^2 = -4ay$,we get $4a = 8$,which implies $a = 2$.
The focus of the parabola $x^2 = -4ay$ is $(0, -a)$,so the focus is $(0, -2)$.
The latus rectum is a line parallel to the $x$-axis passing through the focus $(0, -2)$,given by the equation $y = -2$.
To find the ends of the latus rectum,substitute $y = -2$ into the parabola equation $x^2 = -8y$:
$x^2 = -8(-2) = 16$
$x = \pm 4$.
Thus,the ends of the latus rectum are $(4, -2)$ and $(-4, -2)$.

(D) For the parabola $x^2 = 4ay$,the focus is at $(0, a)$.
The latus rectum is a line segment passing through the focus and perpendicular to the axis of the parabola.
The equation of the line representing the latus rectum is $y = a$.
Substituting $y = a$ into the parabola equation $x^2 = 4ay$,we get $x^2 = 4a(a) = 4a^2$.
Thus,$x = \pm 2a$.
The end points of the latus rectum are $(-2a, a)$ and $(2a, a)$.

(B) Since the axis of the parabola is the $y$-axis,the equation is of the form $x^2 = 4ay$.
Given that the parabola passes through the point $(6, -3)$,we substitute these coordinates into the equation:
$(6)^2 = 4a(-3)$
$36 = -12a$
$a = -3$
Substituting $a = -3$ back into the standard equation $x^2 = 4ay$,we get:
$x^2 = 4(-3)y$
$x^2 = -12y$.

(A) The given equation of the parabola is ${x^2} = -8ay$.
Comparing this with the standard form ${x^2} = -4Ay$,we get $4A = 8a$,which implies $A = 2a$.
The focus of the parabola ${x^2} = -4Ay$ is $(0, -A)$.
Substituting $A = 2a$,the focus is $(0, -2a)$.
The equation of the directrix is $y = A$.
Substituting $A = 2a$,the directrix is $y = 2a$.
Thus,the focus is $(0, -2a)$ and the directrix is $y = 2a$.

(C) By definition of a parabola,the distance from any point $P(x, y)$ on the parabola to the focus $S(3, 0)$ is equal to the perpendicular distance from $P$ to the directrix $x + 3 = 0$.
Let $S = (3, 0)$ and the directrix be $x + 3 = 0$.
Using the definition $SP = PM$,where $PM$ is the perpendicular distance to the directrix:
$SP^2 = PM^2$
$(x - 3)^2 + (y - 0)^2 = (x + 3)^2$
$x^2 - 6x + 9 + y^2 = x^2 + 6x + 9$
$y^2 = 6x + 6x$
$y^2 = 12x$
Thus,the correct option is $C$.

(D) Let the parabola be $y^2 = 4ax$.
Let the focal chord be a line passing through the focus $(a, 0)$.
The equation of a chord with pole $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.
Since this chord passes through the focus $(a, 0)$,we substitute $x = a$ and $y = 0$:
$0 \cdot y_1 = 2a(a + x_1)$
$0 = 2a^2 + 2ax_1$
$2ax_1 = -2a^2$
$x_1 = -a$.
Thus,the locus of the pole $(x_1, y_1)$ is $x = -a$,which is the equation of the directrix.

(A) The equation of the parabola is $y^2 = x$.
Since the power of $y$ is even,replacing $y$ with $-y$ does not change the equation.
Therefore,the parabola is symmetric about the $x$-axis.

(D) Let the point be $(x, y)$.
Given that the ordinate $(y)$ is three times the abscissa $(x)$,we have $y = 3x$.
Substitute $y = 3x$ into the equation of the parabola $y^2 = 18x$:
$(3x)^2 = 18x$
$9x^2 = 18x$
$9x^2 - 18x = 0$
$9x(x - 2) = 0$
This gives $x = 0$ or $x = 2$.
If $x = 0$,then $y = 3(0) = 0$. The point is $(0, 0)$.
If $x = 2$,then $y = 3(2) = 6$. The point is $(2, 6)$.
Comparing with the given options,the correct point is $(2, 6)$.

(D) The distance between the tangent at the vertex $(x + y - 12 = 0)$ and the latus rectum $(x + y - 8 = 0)$ is equal to the distance $a$ between the vertex and the focus.
Using the formula for the distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$,which is $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$:
$a = \frac{|-12 - (-8)|}{\sqrt{1^2 + 1^2}} = \frac{|-4|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
The length of the latus rectum is $4a$.
Length $= 4 \times (2\sqrt{2}) = 8\sqrt{2}$.

(D) The given equation of the parabola is $y^2 + 2y + x = 0$.
Completing the square for the $y$ terms:
$(y^2 + 2y + 1) - 1 + x = 0$
$(y + 1)^2 = -(x - 1)$
Comparing this with the standard form $(y - k)^2 = -4a(x - h)$,we get the vertex $(h, k) = (1, -1)$.
Since the $x$-coordinate is positive and the $y$-coordinate is negative,the point $(1, -1)$ lies in the $IV$ quadrant.

(D) Given the parametric equations:
$x - 2 = t^2$
$y = 2t$
From the second equation,we have $t = \frac{y}{2}$.
Substituting this value of $t$ into the first equation:
$x - 2 = (\frac{y}{2})^2$
$x - 2 = \frac{y^2}{4}$
$y^2 = 4(x - 2)$
Thus,the correct option is $D$.

(A) Given the equation of the parabola: $9x^2 - 6x + 36y + 9 = 0$.
Rearranging the terms to complete the square for $x$: $9x^2 - 6x = -36y - 9$.
Divide by $9$: $x^2 - \frac{2}{3}x = -4y - 1$.
Add $(\frac{1}{3})^2 = \frac{1}{9}$ to both sides: $x^2 - \frac{2}{3}x + \frac{1}{9} = -4y - 1 + \frac{1}{9}$.
$(x - \frac{1}{3})^2 = -4y - \frac{8}{9}$.
$(x - \frac{1}{3})^2 = -4(y + \frac{2}{9})$.
Comparing this with the standard form $(x - h)^2 = 4a(y - k)$,the vertex $(h, k)$ is $(\frac{1}{3}, -\frac{2}{9})$.

(A) The general equation of a parabola with a vertical axis is $y = ax^2 + bx + c$.
Since the parabola passes through $(0, 0)$,we have $0 = a(0)^2 + b(0) + c$,which gives $c = 0$.
Since it passes through $(3, 0)$,we have $0 = a(3)^2 + b(3) + 0$,which simplifies to $9a + 3b = 0$,or $3a + b = 0$,so $b = -3a$.
Since it passes through $(-1, 4)$,we have $4 = a(-1)^2 + b(-1) + 0$,which gives $4 = a - b$.
Substituting $b = -3a$ into the equation $4 = a - b$,we get $4 = a - (-3a) = 4a$,which implies $a = 1$.
Then $b = -3(1) = -3$.
Thus,the equation is $y = x^2 - 3x$,which can be rewritten as $x^2 - 3x - y = 0$.

(A) Since the axis of the parabola is vertical,the equation is of the form $(x - h)^2 = 4a(y - k)$,where $(h, k)$ is the vertex.
Given vertex $(h, k) = (-1, -2)$,the equation becomes $(x + 1)^2 = 4a(y + 2)$.
Since the parabola passes through the point $(3, 6)$,we substitute these coordinates into the equation:
$(3 + 1)^2 = 4a(6 + 2)$
$4^2 = 4a(8)$
$16 = 32a$
$a = \frac{16}{32} = \frac{1}{2}$.
Substituting $a = \frac{1}{2}$ back into the equation:
$(x + 1)^2 = 4(\frac{1}{2})(y + 2)$
$(x + 1)^2 = 2(y + 2)$
$x^2 + 2x + 1 = 2y + 4$
$x^2 + 2x - 2y - 3 = 0$.

(D) Given the equation of the parabola is $x^2 - 4x - 3y + 10 = 0$.
Rearranging the terms,we get $x^2 - 4x = 3y - 10$.
Completing the square on the left side: $x^2 - 4x + 4 = 3y - 10 + 4$.
This simplifies to $(x - 2)^2 = 3y - 6$.
$(x - 2)^2 = 3(y - 2)$.
Comparing this with the standard form $(x - h)^2 = 4a(y - k)$,the axis of the parabola is $x - h = 0$.
Therefore,the axis is $x - 2 = 0$.

(B) Let $P(x, y)$ be any point on the parabola. By the definition of a parabola,the distance from $P$ to the focus $S(-8, -2)$ is equal to the perpendicular distance from $P$ to the directrix $2x - y - 9 = 0$.
$PS = \frac{|2x - y - 9|}{\sqrt{2^2 + (-1)^2}}$
$\sqrt{(x + 8)^2 + (y + 2)^2} = \frac{|2x - y - 9|}{\sqrt{5}}$
Squaring both sides:
$(x + 8)^2 + (y + 2)^2 = \frac{(2x - y - 9)^2}{5}$
$5(x^2 + 16x + 64 + y^2 + 4y + 4) = 4x^2 + y^2 + 81 - 4xy - 36x + 18y$
$5x^2 + 5y^2 + 80x + 20y + 340 = 4x^2 + y^2 - 4xy - 36x + 18y + 81$
$x^2 + 4y^2 + 4xy + 116x + 2y + 259 = 0$.

(A) The focus of the parabola is $S = (-3, 0)$ and the directrix is $x + 5 = 0$.
Since the directrix is a vertical line $(x = -5)$,the axis of the parabola is the $x$-axis $(y = 0)$.
The vertex is the midpoint of the focus and the point where the axis intersects the directrix.
The intersection of the axis $(y = 0)$ and the directrix $(x = -5)$ is $(-5, 0)$.
Vertex $V = \left( \frac{-3 + (-5)}{2}, 0 \right) = (-4, 0)$.
The distance from the vertex to the focus is $a = -3 - (-4) = 1$.
The standard form of a parabola opening to the right is $(y - k)^2 = 4a(x - h)$,where $(h, k)$ is the vertex.
Substituting $h = -4, k = 0$,and $a = 1$,we get $(y - 0)^2 = 4(1)(x - (-4))$.
Thus,the equation is $y^2 = 4(x + 4)$.

(A) The standard form of a parabola with vertex at $(h, k)$ and axis parallel to the $x$-axis is $(y - k)^2 = 4A(x - h)$.
Given the vertex is at $(a, 0)$,we have $h = a$ and $k = 0$.
Thus,the equation is $y^2 = 4A(x - a)$.
The focus is at $(a', 0)$. For a parabola $y^2 = 4A(x - h)$,the focus is at $(h + A, k)$.
Therefore,$a + A = a'$,which gives $A = a' - a$.
Substituting $A$ into the equation,we get $y^2 = 4(a' - a)(x - a)$.

(A) Given equation: $y^2 = 4y - 4x$
Rearranging the terms: $y^2 - 4y = -4x$
Completing the square on the left side: $y^2 - 4y + 4 = -4x + 4$
$(y - 2)^2 = -4(x - 1)$
Comparing this with the standard form $(y - k)^2 = -4a(x - h)$,where the vertex is $(h, k)$ and the focus is $(h - a, k)$:
Here,$h = 1$,$k = 2$,and $4a = 4$,so $a = 1$.
The focus is $(h - a, k) = (1 - 1, 2) = (0, 2)$.

(A) Given the equation of the parabola: $x^2 + 4x + 2y - 7 = 0$
Complete the square for the $x$ terms:
$(x^2 + 4x + 4) - 4 + 2y - 7 = 0$
$(x + 2)^2 + 2y - 11 = 0$
$(x + 2)^2 = -2y + 11$
$(x + 2)^2 = -2(y - 11/2)$
Comparing this with the standard form $(x - h)^2 = 4a(y - k)$,where the vertex is $(h, k)$:
$h = -2$ and $k = 11/2$
Therefore,the vertex is $(-2, 11/2)$.

(D) The general equation of a parabola with a horizontal axis is $x = ay^2 + by + c$.
Since the parabola passes through $(0, 0)$,we have $0 = a(0)^2 + b(0) + c$,which implies $c = 0$.
So,the equation is $x = ay^2 + by$.
Using the point $(0, -1)$,we get $0 = a(-1)^2 + b(-1)$,which means $a - b = 0$,so $a = b$.
Using the point $(6, 1)$,we get $6 = a(1)^2 + b(1)$,which means $a + b = 6$.
Substituting $a = b$ into $a + b = 6$,we get $2a = 6$,so $a = 3$ and $b = 3$.
The equation is $x = 3y^2 + 3y$,or $3y^2 + 3y - x = 0$.
Comparing this with the given options,none of the provided equations match the result.

(C) Given the equation: ${y^2} + 2By + 2Ax + C = 0$
Completing the square for $y$: $(y + B)^2 - B^2 + 2Ax + C = 0$
$(y + B)^2 = -2Ax + B^2 - C$
$(y + B)^2 = -2A(x - \frac{B^2 - C}{2A})$
This is in the form $(y - k)^2 = -4a(x - h)$,where $4a = 2A$,so $a = \frac{A}{2}$.
The vertex is $(h, k) = (\frac{B^2 - C}{2A}, -B)$.
The focus of a parabola $(y - k)^2 = -4a(x - h)$ is $(h - a, k)$.
Focus $= (\frac{B^2 - C}{2A} - \frac{A}{2}, -B) = (\frac{B^2 - C - A^2}{2A}, -B)$.
The equation of the latus rectum is $x = h - a$.
$x = \frac{B^2 - C}{2A} - \frac{A}{2} = \frac{B^2 - C - A^2}{2A}$.

(B) The standard parametric equations of the parabola $y^2 = 4ax$ are given by $x = at^2$ and $y = 2at$.
Comparing the given equation $y^2 = 8x$ with $y^2 = 4ax$,we get $4a = 8$,which implies $a = 2$.
Substituting $a = 2$ into the parametric equations:
$x = 2t^2$
$y = 2(2)t = 4t$
Thus,the parametric equations are $x = 2t^2$ and $y = 4t$.

(B) Given the parametric equations:
$x = \frac{t}{4} \Rightarrow t = 4x$
Substitute $t = 4x$ into the equation for $y$:
$y = \frac{(4x)^2}{4} = \frac{16x^2}{4} = 4x^2$
Rearranging gives $x^2 = \frac{1}{4}y$,which is of the form $x^2 = 4ay$ (where $a = \frac{1}{16}$),representing a parabola.

(C) Given: Vertex $(h, k) = (0, 4)$ and focus $(0, 2)$.
Since the $x$-coordinate of the vertex and focus are the same,the axis of the parabola is vertical.
The distance between the vertex and the focus is $a = |4 - 2| = 2$.
Since the focus is below the vertex,the parabola opens downwards.
The standard equation for a downward-opening parabola is $(x - h)^2 = -4a(y - k)$.
Substituting the values: $(x - 0)^2 = -4(2)(y - 4)$.
$x^2 = -8(y - 4)$.
$x^2 = -8y + 32$.
$x^2 + 8y = 32$.

(D) Given equation: $9x^2 - 6x + 36y + 19 = 0$
Rearranging the terms: $9x^2 - 6x = -36y - 19$
Complete the square for $x$: $(3x - 1)^2 - 1 = -36y - 19$
$(3x - 1)^2 = -36y - 18$
$9(x - 1/3)^2 = -36(y + 1/2)$
$(x - 1/3)^2 = -4(y + 1/2)$
Comparing this with the standard form $(x - h)^2 = -4a(y - k)$,we get $4a = 4$,so $a = 1$.
The length of the latus rectum is $4a = 4$.

(A) Given the equation of the parabola: $9y^2 - 16x - 12y - 57 = 0$
Rearranging the terms to complete the square for $y$:
$9(y^2 - \frac{4}{3}y) = 16x + 57$
$9(y^2 - \frac{4}{3}y + \frac{4}{9}) = 16x + 57 + 4$
$9(y - \frac{2}{3})^2 = 16x + 61$
$(y - \frac{2}{3})^2 = \frac{16}{9}(x + \frac{61}{16})$
This is of the form $(y - k)^2 = 4a(x - h)$,where the axis of the parabola is given by $y - k = 0$.
Here,$k = \frac{2}{3}$,so the axis is $y - \frac{2}{3} = 0$,which simplifies to $3y = 2$.

(B) The standard equation of a parabola with vertex at the origin $(0, 0)$ and axis along the positive $y$-axis is $X^2 = 4AY$,where the length of the latus rectum $l = 4A$,so $A = \frac{l}{4}$.
Substituting $A$,we get $X^2 = lY$.
Given the vertex is at $(a, b)$,we use the transformation $X = x - a$ and $Y = y - b$.
Substituting these into the equation,we get $(x - a)^2 = l(y - b)$.
To match the given options,we can rewrite the right side as $(x - a)^2 = \frac{l}{2}(2y - 2b)$.

(A) The given equation is $y = x^2 - 8x + c$.
We can rewrite this by completing the square:
$y = (x^2 - 8x + 16) - 16 + c$
$y = (x - 4)^2 + (c - 16)$.
The vertex of the parabola $y = a(x - h)^2 + k$ is $(h, k)$.
Comparing,the vertex is $(4, c - 16)$.
Since the vertex lies on the $x$-axis,its $y$-coordinate must be $0$.
Therefore,$c - 16 = 0$,which gives $c = 16$.

(C) Given equation of the parabola is ${y^2} = 5x + 4y + 1$.
Rearranging the terms to complete the square for $y$:
${y^2} - 4y = 5x + 1$.
Adding $4$ to both sides:
${y^2} - 4y + 4 = 5x + 1 + 4$.
${(y - 2)^2} = 5x + 5$.
${(y - 2)^2} = 5(x + 1)$.
Comparing this with the standard form ${(y - k)^2} = 4a(x - h)$,we get $4a = 5$.
Thus,the length of the latus rectum is $5$.