Hyperbola Questions in English

(C) The standard equation of a hyperbola with vertices at $(\pm a, 0)$ is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Given $a = 6$,the equation is $\frac{x^2}{36} - \frac{y^2}{b^2} = 1$.
Since the hyperbola passes through $P(10, 16)$,we have $\frac{100}{36} - \frac{256}{b^2} = 1$.
$\frac{25}{9} - 1 = \frac{256}{b^2} \implies \frac{16}{9} = \frac{256}{b^2} \implies b^2 = \frac{256 \times 9}{16} = 144$.
The equation of the hyperbola is $\frac{x^2}{36} - \frac{y^2}{144} = 1$.
The equation of the normal at $(x_1, y_1)$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$.
Substituting $a^2 = 36, b^2 = 144, x_1 = 10, y_1 = 16$:
$\frac{36x}{10} + \frac{144y}{16} = 36 + 144 = 180$.
$3.6x + 9y = 180$.
Dividing by $1.8$,we get $2x + 5y = 100$.

(D) Differentiating $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with respect to $x$,we get $\frac{2x}{a^{2}}-\frac{2y}{b^{2}}\frac{dy}{dx}=0$.
Thus,$\frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}$.
The slope of the tangent at $(x_{0}, y_{0})$ is $m_{t} = \frac{b^{2}x_{0}}{a^{2}y_{0}}$.
The equation of the tangent is $y-y_{0} = \frac{b^{2}x_{0}}{a^{2}y_{0}}(x-x_{0})$,which simplifies to $\frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}=1$.
The slope of the normal is $m_{n} = -\frac{1}{m_{t}} = -\frac{a^{2}y_{0}}{b^{2}x_{0}}$.
The equation of the normal is $y-y_{0} = -\frac{a^{2}y_{0}}{b^{2}x_{0}}(x-x_{0})$,which can be written as $\frac{y-y_{0}}{a^{2}y_{0}} + \frac{x-x_{0}}{b^{2}x_{0}} = 0$.

(D) Let the midpoint of the chord be $(h, k)$.
The equation of the chord of the circle $x^{2}+y^{2}=25$ with midpoint $(h, k)$ is given by $T=S_1$,which is $xh+yk=h^{2}+k^{2}$.
This can be rewritten as $y = -\frac{h}{k}x + \frac{h^{2}+k^{2}}{k}$.
This line is tangent to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$.
The condition for the line $y=mx+c$ to be a tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2}m^{2}-b^{2}$.
Here,$m=-\frac{h}{k}$ and $c=\frac{h^{2}+k^{2}}{k}$,$a^{2}=9$,$b^{2}=16$.
Substituting these values into the condition: $\left(\frac{h^{2}+k^{2}}{k}\right)^{2} = 9\left(-\frac{h}{k}\right)^{2} - 16$.
$\frac{(h^{2}+k^{2})^{2}}{k^{2}} = \frac{9h^{2}}{k^{2}} - 16$.
$(h^{2}+k^{2})^{2} = 9h^{2} - 16k^{2}$.
Replacing $(h, k)$ with $(x, y)$,the locus is $(x^{2}+y^{2})^{2} = 9x^{2}-16y^{2}$,or $(x^{2}+y^{2})^{2}-9x^{2}+16y^{2}=0$.

(N/A) Comparing the given equation $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ with the standard equation $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,we get:
$a^{2} = 9 \implies a = 3$
$b^{2} = 16 \implies b = 4$
For a hyperbola,$c = \sqrt{a^{2} + b^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$.
$1$. The coordinates of the foci are $(\pm c, 0) = (\pm 5, 0)$.
$2$. The coordinates of the vertices are $(\pm a, 0) = (\pm 3, 0)$.
$3$. The eccentricity $e = \frac{c}{a} = \frac{5}{3}$.
$4$. The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2 \times 16}{3} = \frac{32}{3}$.

(N/A) Dividing the equation by $16$ on both sides,we get $\frac{y^{2}}{16}-\frac{x^{2}}{1}=1$.
Comparing this with the standard equation $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$,we find $a^{2}=16$ and $b^{2}=1$,so $a=4$ and $b=1$.
Calculating $c$: $c = \sqrt{a^{2}+b^{2}} = \sqrt{16+1} = \sqrt{17}$.
The coordinates of the foci are $(0, \pm c) = (0, \pm \sqrt{17})$.
The coordinates of the vertices are $(0, \pm a) = (0, \pm 4)$.
The eccentricity $e = \frac{c}{a} = \frac{\sqrt{17}}{4}$.
The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2(1)}{4} = \frac{1}{2}$.

(A) Since the foci are on the $y$-axis,the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$.
Given the vertices are $(0, \pm \frac{\sqrt{11}}{2})$,we have $a = \frac{\sqrt{11}}{2}$,so $a^{2} = \frac{11}{4}$.
Given the foci are $(0, \pm 3)$,we have $c = 3$,so $c^{2} = 9$.
For a hyperbola,$b^{2} = c^{2} - a^{2} = 9 - \frac{11}{4} = \frac{36 - 11}{4} = \frac{25}{4}$.
Substituting $a^{2}$ and $b^{2}$ into the standard equation:
$\frac{y^{2}}{11/4} - \frac{x^{2}}{25/4} = 1$
$\frac{4y^{2}}{11} - \frac{4x^{2}}{25} = 1$
Multiplying by $275$ (the $LCM$ of $11$ and $25$):
$100y^{2} - 44x^{2} = 275$.

(A) Since the foci are $(0, \pm 12)$,the hyperbola is vertical and $c = 12$.
The length of the latus rectum is given by $\frac{2b^{2}}{a} = 36$,which implies $b^{2} = 18a$.
For a hyperbola,$c^{2} = a^{2} + b^{2}$. Substituting the values,we get $144 = a^{2} + 18a$.
Rearranging gives the quadratic equation $a^{2} + 18a - 144 = 0$.
Factoring the quadratic,we get $(a + 24)(a - 6) = 0$,so $a = -24$ or $a = 6$.
Since $a$ must be positive,we take $a = 6$. Then $b^{2} = 18(6) = 108$.
The equation of the hyperbola is $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$,which is $\frac{y^{2}}{36} - \frac{x^{2}}{108} = 1$.
Multiplying by $108$,we get $3y^{2} - x^{2} = 108$.

(N/A) The given equation is $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$,which can be written as $\frac{x^{2}}{4^{2}}-\frac{y^{2}}{3^{2}}=1$.
Comparing this with the standard equation $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,we get $a=4$ and $b=3$.
We know that $c^{2}=a^{2}+b^{2}$.
Substituting the values,$c^{2}=4^{2}+3^{2}=16+9=25$,so $c=5$.
$1$. The coordinates of the foci are $(\pm c, 0) = (\pm 5, 0)$.
$2$. The coordinates of the vertices are $(\pm a, 0) = (\pm 4, 0)$.
$3$. The eccentricity $e = \frac{c}{a} = \frac{5}{4}$.
$4$. The length of the latus rectum $= \frac{2b^{2}}{a} = \frac{2 \times 9}{4} = \frac{9}{2} = 4.5$.

(N/A) The given equation is $\frac{y^{2}}{9}-\frac{x^{2}}{27}=1$,which can be written as $\frac{y^{2}}{3^{2}}-\frac{x^{2}}{(\sqrt{27})^{2}}=1$.
Comparing this with the standard equation of a vertical hyperbola $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$,we get $a=3$ and $b=\sqrt{27}$.
For a hyperbola,the relation between $a, b,$ and $c$ is $c^{2}=a^{2}+b^{2}$.
Substituting the values,$c^{2}=3^{2}+(\sqrt{27})^{2}=9+27=36$,so $c=6$.
The coordinates of the foci are $(0, \pm c) = (0, \pm 6)$.
The coordinates of the vertices are $(0, \pm a) = (0, \pm 3)$.
The eccentricity $e$ is given by $e=\frac{c}{a}=\frac{6}{3}=2$.
The length of the latus rectum is $\frac{2b^{2}}{a}=\frac{2 \times 27}{3}=18$.

(N/A) The given equation is $9 y^{2}-4 x^{2}=36$.
Dividing both sides by $36$,we get:
$\frac{9 y^{2}}{36} - \frac{4 x^{2}}{36} = 1$
$\frac{y^{2}}{4} - \frac{x^{2}}{9} = 1$
This is the standard form of a hyperbola $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$,where $a^{2} = 4$ and $b^{2} = 9$.
Thus,$a = 2$ and $b = 3$.
For a hyperbola,$c^{2} = a^{2} + b^{2} = 4 + 9 = 13$,so $c = \sqrt{13}$.
$1$. The coordinates of the foci are $(0, \pm c) = (0, \pm \sqrt{13})$.
$2$. The coordinates of the vertices are $(0, \pm a) = (0, \pm 2)$.
$3$. The eccentricity $e = \frac{c}{a} = \frac{\sqrt{13}}{2}$.
$4$. The length of the latus rectum $= \frac{2 b^{2}}{a} = \frac{2 \times 9}{2} = 9$.

(N/A) The given equation is $16 x^{2}-9 y^{2}=576$.
It can be written as:
$\frac{x^{2}}{36}-\frac{y^{2}}{64}=1$
$\Rightarrow \frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}}=1$ $(1)$
On comparing equation $(1)$ with the standard equation of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,we obtain $a=6$ and $b=8$.
We know that $c^{2}=a^{2}+b^{2}$.
$\therefore c^{2}=36+64=100$
$\Rightarrow c=10$.
Therefore:
$1$. The coordinates of the foci are $(\pm 10, 0)$.
$2$. The coordinates of the vertices are $(\pm 6, 0)$.
$3$. Eccentricity,$e = \frac{c}{a} = \frac{10}{6} = \frac{5}{3}$.
$4$. Length of the latus rectum $= \frac{2b^{2}}{a} = \frac{2 \times 64}{6} = \frac{64}{3}$.

The given equation is $5y^{2} - 9x^{2} = 36$.
Dividing by $36$,we get $\frac{y^{2}}{(36/5)} - \frac{x^{2}}{4} = 1$.
Comparing this with the standard equation $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$,we have $a^{2} = \frac{36}{5}$ and $b^{2} = 4$.
Thus,$a = \frac{6}{\sqrt{5}}$ and $b = 2$.
For a hyperbola,$c^{2} = a^{2} + b^{2} = \frac{36}{5} + 4 = \frac{56}{5}$.
So,$c = \sqrt{\frac{56}{5}} = \frac{2\sqrt{14}}{\sqrt{5}}$.
Coordinates of the foci are $(0, \pm c) = (0, \pm \frac{2\sqrt{14}}{\sqrt{5}})$.
Coordinates of the vertices are $(0, \pm a) = (0, \pm \frac{6}{\sqrt{5}})$.
Eccentricity $e = \frac{c}{a} = \frac{2\sqrt{14}/\sqrt{5}}{6/\sqrt{5}} = \frac{\sqrt{14}}{3}$.
Length of the latus rectum $= \frac{2b^{2}}{a} = \frac{2 \times 4}{6/\sqrt{5}} = \frac{4\sqrt{5}}{3}$.

(N/A) The given equation is $49 y^{2}-16 x^{2}=784$.
Dividing both sides by $784$,we get:
$\frac{49 y^{2}}{784} - \frac{16 x^{2}}{784} = 1$
$\frac{y^{2}}{16} - \frac{x^{2}}{49} = 1$
$\frac{y^{2}}{4^{2}} - \frac{x^{2}}{7^{2}} = 1$ $(1)$
Comparing equation $(1)$ with the standard equation of the hyperbola $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$,we get $a = 4$ and $b = 7$.
We know that $c^{2} = a^{2} + b^{2}$.
$c^{2} = 16 + 49 = 65$
$c = \sqrt{65}$
Coordinates of the foci are $(0, \pm \sqrt{65})$.
Coordinates of the vertices are $(0, \pm 4)$.
Eccentricity $e = \frac{c}{a} = \frac{\sqrt{65}}{4}$.
Length of the latus rectum $= \frac{2 b^{2}}{a} = \frac{2 \times 49}{4} = \frac{49}{2} = 24.5$.

(A) The vertices are $(\pm 2, 0)$,which lie on the $x$-axis.
Thus,the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.
Given vertices $(\pm a, 0) = (\pm 2, 0)$,we have $a = 2$,so $a^{2} = 4$.
Given foci $(\pm c, 0) = (\pm 3, 0)$,we have $c = 3$,so $c^{2} = 9$.
For a hyperbola,we know the relation $c^{2} = a^{2} + b^{2}$.
Substituting the values,$9 = 4 + b^{2}$,which gives $b^{2} = 5$.
Therefore,the equation of the hyperbola is $\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1$.

(A) The vertices are $(0, \pm 5)$,which lie on the $y$-axis.
Thus,the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$.
Given the vertices $(0, \pm a) = (0, \pm 5)$,we have $a = 5$,so $a^{2} = 25$.
Given the foci $(0, \pm c) = (0, \pm 8)$,we have $c = 8$,so $c^{2} = 64$.
For a hyperbola,we know the relation $c^{2} = a^{2} + b^{2}$.
Substituting the values,$64 = 25 + b^{2}$,which gives $b^{2} = 64 - 25 = 39$.
Substituting $a^{2}$ and $b^{2}$ into the standard form,the equation is $\frac{y^{2}}{25} - \frac{x^{2}}{39} = 1$.

(A) The vertices are $(0, \pm 3)$,which implies the hyperbola is along the $y$-axis.
The standard equation is $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$.
Given vertices $(0, \pm a) = (0, \pm 3)$,we have $a = 3$,so $a^{2} = 9$.
Given foci $(0, \pm c) = (0, \pm 5)$,we have $c = 5$,so $c^{2} = 25$.
For a hyperbola,$c^{2} = a^{2} + b^{2}$.
Substituting the values: $25 = 9 + b^{2}$.
$b^{2} = 25 - 9 = 16$.
Thus,the equation is $\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$.

(A) The foci are $(\pm 5, 0)$,which lie on the $x$-axis. The standard form of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.
Given the foci $(\pm c, 0) = (\pm 5, 0)$,we have $c = 5$.
The length of the transverse axis is $2a = 8$,which gives $a = 4$.
Using the relation $c^{2} = a^{2} + b^{2}$,we get $5^{2} = 4^{2} + b^{2}$.
$25 = 16 + b^{2} \Rightarrow b^{2} = 25 - 16 = 9$.
Substituting $a^{2} = 16$ and $b^{2} = 9$ into the standard equation,we get $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$.

(A) Given the foci are $(0, \pm 13)$,the hyperbola is vertical and centered at the origin. The standard form is $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$.
Since the foci are $(0, \pm c)$,we have $c = 13$.
The length of the conjugate axis is $2b = 24$,which implies $b = 12$.
Using the relation $c^{2} = a^{2} + b^{2}$,we get $13^{2} = a^{2} + 12^{2}$.
$169 = a^{2} + 144 \Rightarrow a^{2} = 169 - 144 = 25$.
Substituting $a^{2} = 25$ and $b^{2} = 144$ into the standard equation,we get $\frac{y^{2}}{25} - \frac{x^{2}}{144} = 1$.

(A) Given,the foci are $(\pm 3 \sqrt{5}, 0)$. Since the foci lie on the $x$-axis,the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.
Here,$c = 3 \sqrt{5}$,so $c^{2} = (3 \sqrt{5})^{2} = 9 \times 5 = 45$.
The length of the latus rectum is given as $8$,so $\frac{2b^{2}}{a} = 8$,which implies $b^{2} = 4a$.
We know the relation $c^{2} = a^{2} + b^{2}$. Substituting the values,we get $a^{2} + 4a = 45$,or $a^{2} + 4a - 45 = 0$.
Factoring the quadratic equation: $(a + 9)(a - 5) = 0$. Since $a > 0$,we have $a = 5$.
Then $b^{2} = 4a = 4 \times 5 = 20$.
Substituting $a^{2} = 25$ and $b^{2} = 20$ into the standard equation,we get $\frac{x^{2}}{25} - \frac{y^{2}}{20} = 1$.

(A) The foci are $(\pm 4, 0)$,which lie on the $x$-axis. The standard equation of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.
Given the foci are $(\pm c, 0)$,we have $c = 4$,so $c^{2} = 16$.
The length of the latus rectum is given by $\frac{2b^{2}}{a} = 12$,which implies $b^{2} = 6a$.
Using the relation $c^{2} = a^{2} + b^{2}$,we substitute $c^{2} = 16$ and $b^{2} = 6a$:
$a^{2} + 6a = 16$
$a^{2} + 6a - 16 = 0$
$(a + 8)(a - 2) = 0$
Since $a > 0$,we have $a = 2$.
Then $b^{2} = 6(2) = 12$.
Substituting $a^{2} = 4$ and $b^{2} = 12$ into the standard equation,we get $\frac{x^{2}}{4} - \frac{y^{2}}{12} = 1$.

(A) The vertices are $(\pm 7, 0)$,which lie on the $x$-axis. Thus,the equation is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Given $a = 7$ and eccentricity $e = \frac{c}{a} = \frac{4}{3}$.
Substituting $a = 7$,we get $\frac{c}{7} = \frac{4}{3}$,so $c = \frac{28}{3}$.
Using the relation $c^2 = a^2 + b^2$,we have $b^2 = c^2 - a^2 = (\frac{28}{3})^2 - 7^2 = \frac{784}{9} - 49 = \frac{784 - 441}{9} = \frac{343}{9}$.
Substituting $a^2 = 49$ and $b^2 = \frac{343}{9}$ into the standard equation,we get $\frac{x^2}{49} - \frac{y^2}{343/9} = 1$,which simplifies to $\frac{x^2}{49} - \frac{9y^2}{343} = 1$.

(A) The foci are $(0, \pm \sqrt{10})$,which lie on the $y$-axis.
Thus,the equation is of the form $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$.
Here,$c = \sqrt{10}$,so $c^{2} = 10$.
We know $c^{2} = a^{2} + b^{2}$,so $a^{2} + b^{2} = 10$,which means $b^{2} = 10 - a^{2}$.
The hyperbola passes through $(2, 3)$,so $\frac{3^{2}}{a^{2}} - \frac{2^{2}}{b^{2}} = 1$,or $\frac{9}{a^{2}} - \frac{4}{b^{2}} = 1$.
Substituting $b^{2} = 10 - a^{2}$: $\frac{9}{a^{2}} - \frac{4}{10 - a^{2}} = 1$.
$9(10 - a^{2}) - 4a^{2} = a^{2}(10 - a^{2})$.
$90 - 9a^{2} - 4a^{2} = 10a^{2} - a^{4}$.
$a^{4} - 23a^{2} + 90 = 0$.
$(a^{2} - 18)(a^{2} - 5) = 0$.
Since $c^{2} > a^{2}$,$10 > a^{2}$,so $a^{2} = 5$.
Then $b^{2} = 10 - 5 = 5$.
The equation is $\frac{y^{2}}{5} - \frac{x^{2}}{5} = 1$.

(B) The slope of the line $2x - y = 0$ is $2$. Since the tangent is parallel to this line,its slope is $m = 2$.
The equation of the tangent to the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{2} = 1$ at the point $(x_{1}, y_{1})$ is given by $\frac{xx_{1}}{4} - \frac{yy_{1}}{2} = 1$.
The slope of this tangent is $\frac{x_{1}/4}{y_{1}/2} = \frac{x_{1}}{2y_{1}}$.
Equating the slopes: $\frac{x_{1}}{2y_{1}} = 2 \Rightarrow x_{1} = 4y_{1} \quad (1)$.
Since $(x_{1}, y_{1})$ lies on the hyperbola,we have $\frac{x_{1}^{2}}{4} - \frac{y_{1}^{2}}{2} = 1 \quad (2)$.
Substituting $(1)$ into $(2)$: $\frac{(4y_{1})^{2}}{4} - \frac{y_{1}^{2}}{2} = 1 \Rightarrow 4y_{1}^{2} - \frac{y_{1}^{2}}{2} = 1$.
$\frac{7y_{1}^{2}}{2} = 1 \Rightarrow y_{1}^{2} = \frac{2}{7}$.
Now,calculate $x_{1}^{2} + 5y_{1}^{2} = (4y_{1})^{2} + 5y_{1}^{2} = 16y_{1}^{2} + 5y_{1}^{2} = 21y_{1}^{2}$.
Substituting $y_{1}^{2} = \frac{2}{7}$: $21 \times \frac{2}{7} = 3 \times 2 = 6$.

(B) The given 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 eccentricity $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$.
The foci are $(\pm ae, 0) = (\pm 2 \times \frac{1}{2}, 0) = (\pm 1, 0)$.
For the hyperbola,the length of the transverse axis is $2a_{h} = \sqrt{2}$,so $a_{h} = \frac{1}{\sqrt{2}}$ and $a_{h}^{2} = \frac{1}{2}$.
Let the hyperbola be $\frac{x^{2}}{a_{h}^{2}} - \frac{y^{2}}{b_{h}^{2}} = 1$,i.e.,$\frac{x^{2}}{1/2} - \frac{y^{2}}{b_{h}^{2}} = 1$.
The foci of the hyperbola are $(\pm a_{h}e_{h}, 0)$,where $e_{h} = \sqrt{1 + \frac{b_{h}^{2}}{a_{h}^{2}}} = \sqrt{1 + 2b_{h}^{2}}$.
Thus,the foci are $(\pm \sqrt{a_{h}^{2} + b_{h}^{2}}, 0) = (\pm \sqrt{\frac{1}{2} + b_{h}^{2}}, 0)$.
Since the foci are the same,$\sqrt{\frac{1}{2} + b_{h}^{2}} = 1$,which implies $\frac{1}{2} + b_{h}^{2} = 1$,so $b_{h}^{2} = \frac{1}{2}$.
The equation of the hyperbola is $\frac{x^{2}}{1/2} - \frac{y^{2}}{1/2} = 1$,or $x^{2} - y^{2} = \frac{1}{2}$.
Checking the points:
For option $B$: $(\sqrt{3/2})^{2} - (1/\sqrt{2})^{2} = \frac{3}{2} - \frac{1}{2} = 1 \neq \frac{1}{2}$.
Thus,the point $\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}\right)$ does not lie on the hyperbola.

(A) Since the point $(3,3)$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,we have $\frac{9}{a^{2}}-\frac{9}{b^{2}}=1$ $(i)$.
The equation of the normal at $(x_{1}, y_{1})$ to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $\frac{a^{2}x}{x_{1}} + \frac{b^{2}y}{y_{1}} = a^{2} + b^{2}$.
Substituting $(x_{1}, y_{1}) = (3,3)$,the normal is $\frac{a^{2}x}{3} + \frac{b^{2}y}{3} = a^{2} + b^{2}$.
This normal passes through $(9,0)$,so $\frac{a^{2}(9)}{3} + 0 = a^{2} + b^{2}$ $\Rightarrow 3a^{2} = a^{2} + b^{2}$ $\Rightarrow b^{2} = 2a^{2}$ $(ii)$.
Substituting $(ii)$ into $(i)$: $\frac{9}{a^{2}} - \frac{9}{2a^{2}} = 1$ $\Rightarrow \frac{18-9}{2a^{2}} = 1$ $\Rightarrow 9 = 2a^{2}$ $\Rightarrow a^{2} = \frac{9}{2}$.
Then $b^{2} = 2(\frac{9}{2}) = 9$.
The eccentricity $e$ is given by $e^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{9}{9/2} = 1 + 2 = 3$.
Thus,the ordered pair $(a^{2}, e^{2})$ is $(\frac{9}{2}, 3)$.

(B) The condition for the line $y=mx+c$ to be a tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2}m^{2}-b^{2}$.
For the given hyperbola $\frac{x^{2}}{100}-\frac{y^{2}}{64}=1$,we have $a^{2}=100$ and $b^{2}=64$,so $c^{2}=100m^{2}-64$.
The condition for the line $y=mx+c$ to be a tangent to the circle $x^{2}+y^{2}=r^{2}$ is $c^{2}=r^{2}(1+m^{2})$.
For the given circle $x^{2}+y^{2}=36$,we have $r^{2}=36$,so $c^{2}=36(1+m^{2})$.
Equating the two expressions for $c^{2}$:
$100m^{2}-64=36(1+m^{2})$
$100m^{2}-64=36+36m^{2}$
$64m^{2}=100$
$m^{2}=\frac{100}{64}=\frac{25}{16}$.
Now,substitute $m^{2}$ back into the circle's condition:
$c^{2}=36(1+\frac{25}{16})$
$c^{2}=36(\frac{16+25}{16})$
$c^{2}=36(\frac{41}{16})$
$c^{2}=\frac{9 \times 41}{4}$
$4c^{2}=369$.

(D) The curve is $x^{2}y^{2}=1$,which implies $xy=1$ or $xy=-1$. Let the vertices be $A(t_1, 1/t_1)$,$B(t_2, -1/t_2)$,$C(-t_1, -1/t_1)$,and $D(-t_2, 1/t_2)$.
Since $ABCD$ is a square,the midpoint of $AB$ must lie on the curve $xy=1$ or $xy=-1$. The midpoint of $AB$ is $M = (\frac{t_1+t_2}{2}, \frac{1/t_1 - 1/t_2}{2})$.
For $M$ to lie on $xy=1$,we have $(\frac{t_1+t_2}{2})(\frac{t_2-t_1}{2t_1t_2}) = 1$,which simplifies to $t_2^2 - t_1^2 = 4t_1t_2$.
Also,the slope of $AB$ must be $-1$ (since $ABCD$ is a square and the vertices are symmetric about the origin). The slope of $AB$ is $\frac{-1/t_2 - 1/t_1}{t_2 - t_1} = \frac{-(t_1+t_2)}{t_1t_2(t_2-t_1)} = -1$,which implies $t_1+t_2 = t_1t_2(t_2-t_1)$.
Solving these equations,we find $t_1t_2 = 1$ and $t_2^2 - t_1^2 = 4$. Thus $t_2^2 + t_1^2 = \sqrt{4^2 + 4(1)^2} = \sqrt{20} = 2\sqrt{5}$.
The side length squared is $AB^2 = (t_2-t_1)^2 + (-1/t_2 - 1/t_1)^2 = (t_2-t_1)^2 + (\frac{t_1+t_2}{t_1t_2})^2 = (t_2^2+t_1^2 - 2t_1t_2) + (t_1+t_2)^2 = (2\sqrt{5}-2) + (2\sqrt{5}+2) = 4\sqrt{5}$.
Area of $ABCD = AB^2 = 4\sqrt{5}$.
Therefore,the square of the area is $(4\sqrt{5})^2 = 16 \times 5 = 80$.

(C) The given hyperbola is $x^{2}-2y^{2}=4$,which can be written as $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$.
Here,$a^{2}=4$ and $b^{2}=2$.
The eccentricity $e$ is given by $e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{2}{4}}=\sqrt{\frac{3}{2}}$.
The focus $F$ is $(ae, 0) = (2 \cdot \sqrt{\frac{3}{2}}, 0) = (\sqrt{6}, 0)$.
The equation of the tangent at $P(4, \sqrt{6})$ to the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$ is $\frac{x x_{1}}{4}-\frac{y y_{1}}{2}=1$.
Substituting $(x_{1}, y_{1}) = (4, \sqrt{6})$,we get $\frac{4x}{4}-\frac{\sqrt{6}y}{2}=1$,which simplifies to $x - \frac{\sqrt{6}}{2}y = 1$,or $2x - y\sqrt{6} = 2$.
To find $Q$,set $y=0$ in the tangent equation: $2x=2 \Rightarrow x=1$. So,$Q(1, 0)$.
To find $R$,substitute $x=\sqrt{6}$ (the latus rectum) into the tangent equation: $2(\sqrt{6}) - y\sqrt{6} = 2 \Rightarrow y\sqrt{6} = 2\sqrt{6}-2 \Rightarrow y = 2 - \frac{2}{\sqrt{6}}$.
Thus,$R(\sqrt{6}, 2 - \frac{2}{\sqrt{6}})$.
The area of $\Delta QFR$ with vertices $Q(1, 0)$,$F(\sqrt{6}, 0)$,and $R(\sqrt{6}, 2 - \frac{2}{\sqrt{6}})$ is $\frac{1}{2} \times \text{base} \times \text{height}$.
Base $QF = |\sqrt{6}-1|$.
Height $FR = |2 - \frac{2}{\sqrt{6}}|$.
Area $= \frac{1}{2} \times (\sqrt{6}-1) \times 2(1 - \frac{1}{\sqrt{6}}) = (\sqrt{6}-1) \times \frac{\sqrt{6}-1}{\sqrt{6}} = \frac{(\sqrt{6}-1)^{2}}{\sqrt{6}} = \frac{6+1-2\sqrt{6}}{\sqrt{6}} = \frac{7}{\sqrt{6}}-2$.

(B) Given lines are $(\sqrt{3})kx + ky = 4\sqrt{3}$ and $\sqrt{3}x - y = 4\sqrt{3}k$.
From the first equation,$k = \frac{4\sqrt{3}}{\sqrt{3}x + y}$.
From the second equation,$k = \frac{\sqrt{3}x - y}{4\sqrt{3}}$.
Equating the two expressions for $k$:
$\frac{4\sqrt{3}}{\sqrt{3}x + y} = \frac{\sqrt{3}x - y}{4\sqrt{3}}$
$(\sqrt{3}x - y)(\sqrt{3}x + y) = 16(3) = 48$
$3x^2 - y^2 = 48$
Dividing by $48$,we get $\frac{x^2}{16} - \frac{y^2}{48} = 1$.
This is the equation of a hyperbola with $a^2 = 16$ and $b^2 = 48$.
The eccentricity $e$ is given by $b^2 = a^2(e^2 - 1)$.
$48 = 16(e^2 - 1)$
$3 = e^2 - 1$
$e^2 = 4$
$e = 2$.

(B) For the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$,we have $a^{2}=25$ and $b^{2}=16$.
The eccentricity $e_{1} = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
The foci of the ellipse are $(\pm ae_{1}, 0) = (\pm 5 \times \frac{3}{5}, 0) = (\pm 3, 0)$.
Given that the product of the eccentricities of the ellipse and the hyperbola is $1$,we have $e_{1} \times e_{2} = 1$ $\Rightarrow \frac{3}{5} \times e_{2} = 1$ $\Rightarrow e_{2} = \frac{5}{3}$.
Let the equation of the hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
Since the hyperbola passes through the foci of the ellipse $(\pm 3, 0)$,we have $\frac{3^{2}}{a^{2}} = 1 \Rightarrow a^{2} = 9$.
For a hyperbola,$b^{2} = a^{2}(e_{2}^{2}-1) = 9((\frac{5}{3})^{2}-1) = 9(\frac{25}{9}-1) = 9(\frac{16}{9}) = 16$.
Thus,the equation of the hyperbola is $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$.

(C) Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and eccentricity $e = \frac{\sqrt{5}}{2}$.
We know $b^{2} = a^{2}(e^{2}-1) = a^{2}(\frac{5}{4}-1) = \frac{a^{2}}{4}$,so $a^{2} = 4b^{2}$.
Since $P(-2 \sqrt{6}, \sqrt{3})$ lies on the hyperbola,$\frac{(-2 \sqrt{6})^{2}}{4b^{2}} - \frac{(\sqrt{3})^{2}}{b^{2}} = 1$.
$\frac{24}{4b^{2}} - \frac{3}{b^{2}} = 1 \Rightarrow \frac{6}{b^{2}} - \frac{3}{b^{2}} = 1 \Rightarrow \frac{3}{b^{2}} = 1 \Rightarrow b^{2} = 3$,so $b = \sqrt{3}$ and $a^{2} = 12$,so $a = 2 \sqrt{3}$.
The hyperbola is $\frac{x^{2}}{12} - \frac{y^{2}}{3} = 1$.
The tangent at $P(x_{1}, y_{1})$ is $\frac{xx_{1}}{a^{2}} - \frac{yy_{1}}{b^{2}} = 1$.
$\frac{x(-2 \sqrt{6})}{12} - \frac{y(\sqrt{3})}{3} = 1 \Rightarrow -\frac{x \sqrt{6}}{6} - \frac{y}{\sqrt{3}} = 1$.
For the conjugate axis,set $x = 0$: $-\frac{y}{\sqrt{3}} = 1 \Rightarrow y = -\sqrt{3}$. Thus $Q = (0, -\sqrt{3})$.
The slope of the tangent is $m = \frac{b^{2}x_{1}}{a^{2}y_{1}} = \frac{3(-2 \sqrt{6})}{12(\sqrt{3})} = \frac{-6 \sqrt{6}}{12 \sqrt{3}} = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$.
The slope of the normal is $m' = -\frac{1}{m} = \sqrt{2}$.
The equation of the normal at $P$ is $y - \sqrt{3} = \sqrt{2}(x + 2 \sqrt{6})$.
For the conjugate axis,set $x = 0$: $y - \sqrt{3} = \sqrt{2}(2 \sqrt{6}) = 2 \sqrt{12} = 4 \sqrt{3}$.
$y = 4 \sqrt{3} + \sqrt{3} = 5 \sqrt{3}$. Thus $R = (0, 5 \sqrt{3})$.
The distance $QR = |5 \sqrt{3} - (-\sqrt{3})| = |6 \sqrt{3}| = 6 \sqrt{3}$.

(D) The equation of the hyperbola is $2x^2 - y^2 = 2$,which can be written as $\frac{x^2}{1} - \frac{y^2}{2} = 1$. Here $a^2 = 1$ and $b^2 = 2$.
The normal at point $(x_1, y_1)$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$.
For point $A(\sec \theta, 2 \tan \theta)$,the normal is $\frac{1 \cdot x}{\sec \theta} + \frac{2 \cdot y}{2 \tan \theta} = 1 + 2 = 3$,which simplifies to $x \cos \theta + y \cot \theta = 3$.
For point $B(\sec \phi, 2 \tan \phi)$,the normal is $x \cos \phi + y \cot \phi = 3$.
Given $\phi = \frac{\pi}{2} - \theta$,we have $\cos \phi = \sin \theta$ and $\cot \phi = \tan \theta$.
So the equations are:
$1) x \cos \theta + y \cot \theta = 3$
$2) x \sin \theta + y \tan \theta = 3$
Solving for $y = \beta$ by eliminating $x$: Multiply $(1)$ by $\sin \theta$ and $(2)$ by $\cos \theta$:
$x \cos \theta \sin \theta + y \cot \theta \sin \theta = 3 \sin \theta$
$x \sin \theta \cos \theta + y \tan \theta \cos \theta = 3 \cos \theta$
Subtracting the two equations:
$y(\cos \theta - \sin \theta) = 3(\sin \theta - \cos \theta)$
$y = -3$
Thus,$\beta = -3$. Then $(2\beta)^2 = (2 \times -3)^2 = (-6)^2 = 36$.

(B) The given hyperbola is $16(x+1)^{2}-9(y-2)^{2}=164+16-36=144$.
This simplifies to $\frac{(x+1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1$.
The center of the hyperbola is $(-1, 2)$.
Here $a^{2}=9$ and $b^{2}=16$,so $e=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$.
The foci are $(h \pm ae, k) = (-1 \pm 3 \times \frac{5}{3}, 2) = (-1 \pm 5, 2)$,which are $(4, 2)$ and $(-6, 2)$.
Let $P(\alpha, \beta)$ be a point on the hyperbola.
Let $G(x, y)$ be the centroid of the triangle formed by $P(\alpha, \beta)$,$(4, 2)$,and $(-6, 2)$.
Then $x=\frac{\alpha+4-6}{3} = \frac{\alpha-2}{3} \Rightarrow \alpha=3x+2$.
And $y=\frac{\beta+2+2}{3} = \frac{\beta+4}{3} \Rightarrow \beta=3y-4$.
Since $P(\alpha, \beta)$ lies on the hyperbola $16(x+1)^{2}-9(y-2)^{2}=144$,we substitute $\alpha$ and $\beta$:
$16(3x+2+1)^{2}-9(3y-4-2)^{2}=144$
$16(3x+3)^{2}-9(3y-6)^{2}=144$
$16 \times 9(x+1)^{2}-9 \times 9(y-2)^{2}=144$
$144(x+1)^{2}-81(y-2)^{2}=144$
Divide by $9$:
$16(x+1)^{2}-9(y-2)^{2}=16$
$16(x^{2}+2x+1)-9(y^{2}-4y+4)=16$
$16x^{2}+32x+16-9y^{2}+36y-36=16$
$16x^{2}-9y^{2}+32x+36y-36=0$.

(D) The given hyperbola is $a^{2}x^{2} - y^{2} = b^{2}$,which can be written as $\frac{x^{2}}{(b/a)^{2}} - \frac{y^{2}}{b^{2}} = 1$.
For a line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$,the condition is $c^{2} = A^{2}m^{2} - B^{2}$.
Here,the line is $\lambda x - 2y = \mu$,which can be rewritten as $y = \frac{\lambda}{2}x - \frac{\mu}{2}$.
Comparing with $y = mx + c$,we have $m = \frac{\lambda}{2}$ and $c = -\frac{\mu}{2}$.
Here $A^{2} = \frac{b^{2}}{a^{2}}$ and $B^{2} = b^{2}$.
Substituting these into the condition $c^{2} = A^{2}m^{2} - B^{2}$:
$(-\frac{\mu}{2})^{2} = \frac{b^{2}}{a^{2}}(\frac{\lambda}{2})^{2} - b^{2}$
$\frac{\mu^{2}}{4} = \frac{b^{2}\lambda^{2}}{4a^{2}} - b^{2}$
Multiply by $\frac{4}{b^{2}}$:
$\frac{\mu^{2}}{b^{2}} = \frac{\lambda^{2}}{a^{2}} - 4$
Rearranging the terms,we get $\frac{\lambda^{2}}{a^{2}} - \frac{\mu^{2}}{b^{2}} = 4$.

(B) Given eccentricity $e = \frac{5}{4}$,so $e^{2} = 1 + \frac{b^{2}}{a^{2}} = \frac{25}{16}$ $\Rightarrow \frac{b^{2}}{a^{2}} = \frac{9}{16}$ $\Rightarrow b^{2} = \frac{9}{16}a^{2}$.
The point $\left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$ lies on the hyperbola,so $\frac{64}{5a^{2}} - \frac{144}{25b^{2}} = 1$.
Substituting $b^{2} = \frac{9}{16}a^{2}$,we get $\frac{64}{5a^{2}} - \frac{144}{25 \times (9/16)a^{2}} = 1$ $\Rightarrow \frac{64}{5a^{2}} - \frac{144 \times 16}{225a^{2}} = 1$ $\Rightarrow \frac{64}{5a^{2}} - \frac{256}{25a^{2}} = 1$.
$\frac{320 - 256}{25a^{2}} = 1$ $\Rightarrow \frac{64}{25a^{2}} = 1$ $\Rightarrow a^{2} = \frac{64}{25}$ $\Rightarrow a = \frac{8}{5}$.
Then $b^{2} = \frac{9}{16} \times \frac{64}{25} = \frac{9 \times 4}{25} = \frac{36}{25} \Rightarrow b = \frac{6}{5}$.
The equation of the normal at $(x_{1}, y_{1})$ is $\frac{a^{2}x}{x_{1}} + \frac{b^{2}y}{y_{1}} = a^{2} + b^{2}$.
Substituting values: $\frac{(64/25)x}{8/\sqrt{5}} + \frac{(36/25)y}{12/5} = \frac{64}{25} + \frac{36}{25} = \frac{100}{25} = 4$.
$\frac{8\sqrt{5}}{25}x + \frac{3}{5}y = 4 \Rightarrow 8\sqrt{5}x + 15y = 100$.
Comparing with $8\sqrt{5}x + \beta y = \lambda$,we get $\beta = 15$ and $\lambda = 100$.
Thus,$\lambda - \beta = 100 - 15 = 85$.

(C) Given the hyperbola equation $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$.
Since the point $(8, 3\sqrt{3})$ lies on the hyperbola,we substitute the coordinates into the equation:
$\frac{64}{a^{2}}-\frac{(3\sqrt{3})^{2}}{9}=1$
$\frac{64}{a^{2}}-\frac{27}{9}=1$
$\frac{64}{a^{2}}-3=1$ $\Rightarrow \frac{64}{a^{2}}=4$ $\Rightarrow a^{2}=16$.
The equation of the normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at point $(x_{1}, y_{1})$ is given by $\frac{a^{2}x}{x_{1}}+\frac{b^{2}y}{y_{1}}=a^{2}+b^{2}$.
Substituting $a^{2}=16, b^{2}=9, x_{1}=8, y_{1}=3\sqrt{3}$:
$\frac{16x}{8}+\frac{9y}{3\sqrt{3}}=16+9$
$2x+\sqrt{3}y=25$.
Checking the options:
For option $A$: $2(15)+\sqrt{3}(-2\sqrt{3}) = 30-6 = 24 \neq 25$.
For option $B$: $2(9)+\sqrt{3}(2\sqrt{3}) = 18+6 = 24 \neq 25$.
For option $C$: $2(-1)+\sqrt{3}(9\sqrt{3}) = -2+27 = 25$.
Thus,the normal passes through the point $(-1, 9\sqrt{3})$.

(B) The equation of the tangent $L_{1}$ to the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{4}=1$ at point $(4 \sec \theta, 2 \tan \theta)$ is $\frac{x \sec \theta}{4}-\frac{y \tan \theta}{2}=1$.
The slope of $L_{1}$ is $m_{1} = \frac{\sec \theta}{2 \tan \theta}$.
Let the point of intersection be $(h, k)$. Since $L_{2}$ passes through $(0, 0)$ and $(h, k)$,its slope is $m_{2} = \frac{k}{h}$.
Since $L_{1} \perp L_{2}$,we have $m_{1} m_{2} = -1$,so $\frac{k}{h} \cdot \frac{\sec \theta}{2 \tan \theta} = -1$,which gives $\frac{k}{2h \sin \theta} = -1$,or $\sin \theta = -\frac{k}{2h}$.
Then $\cos^{2} \theta = 1 - \sin^{2} \theta = 1 - \frac{k^{2}}{4h^{2}} = \frac{4h^{2}-k^{2}}{4h^{2}}$,so $\cos \theta = \frac{\sqrt{4h^{2}-k^{2}}}{2h}$.
Substituting these into the tangent equation $\frac{h \sec \theta}{4} - \frac{k \tan \theta}{2} = 1$:
$\frac{h}{4} \cdot \frac{2h}{\sqrt{4h^{2}-k^{2}}} - \frac{k}{2} \cdot \left( \frac{-k}{\sqrt{4h^{2}-k^{2}}} \right) = 1$
$\frac{2h^{2}+k^{2}}{2\sqrt{4h^{2}-k^{2}}} = 1 \implies 2h^{2}+k^{2} = 2\sqrt{4h^{2}-k^{2}}$
Squaring both sides: $(2h^{2}+k^{2})^{2} = 4(4h^{2}-k^{2}) = 16h^{2}-4k^{2}$.
Replacing $(h, k)$ with $(x, y)$,we get $(2x^{2}+y^{2})^{2} = 16x^{2}-4y^{2}$.
Wait,the standard locus of the foot of the perpendicular from the origin to a tangent of a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $(x^{2}+y^{2})^{2} = a^{2}x^{2}-b^{2}y^{2}$.
Here $a^{2}=16$ and $b^{2}=4$,so the locus is $(x^{2}+y^{2})^{2} = 16x^{2}-4y^{2}$.
Thus $\alpha=16$ and $\beta=-4$.
$\alpha+\beta = 16-4 = 12$.

(D) For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,we have $e=\sqrt{1+\frac{b^{2}}{a^{2}}}$ and $\ell=\frac{2b^{2}}{a}$.
Given $e^{2}=\frac{11}{14}\ell$,we get $1+\frac{b^{2}}{a^{2}}=\frac{11}{14} \cdot \frac{2b^{2}}{a}$,which simplifies to $\frac{a^{2}+b^{2}}{a^{2}}=\frac{11b^{2}}{7a} \dots (1)$.
For the conjugate hyperbola $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$,we have $e^{\prime}=\sqrt{1+\frac{a^{2}}{b^{2}}}$ and $\ell^{\prime}=\frac{2a^{2}}{b}$.
Given $(e^{\prime})^{2}=\frac{11}{8}\ell^{\prime}$,we get $1+\frac{a^{2}}{b^{2}}=\frac{11}{8} \cdot \frac{2a^{2}}{b}$,which simplifies to $\frac{a^{2}+b^{2}}{b^{2}}=\frac{11a^{2}}{4b} \dots (2)$.
Dividing $(1)$ by $(2)$,we get $\frac{b^{2}}{a^{2}}=\frac{11b^{2}}{7a} \cdot \frac{4b}{11a^{2}} = \frac{4b^{3}}{7a^{3}}$.
Thus,$7a^{3}=4b^{3} \cdot \frac{a^{2}}{b^{2}} \implies 7a=4b \dots (3)$.
Substituting $a=\frac{4b}{7}$ into $(2)$: $\frac{(\frac{16b^{2}}{49})+b^{2}}{b^{2}} = \frac{11}{4} \cdot \frac{16b^{2}}{49b} \implies \frac{65}{49} = \frac{11 \cdot 4b}{49} \implies 65 = 44b \implies b = \frac{65}{44}$.
Then $a = \frac{4}{7} \cdot \frac{65}{44} = \frac{65}{77}$.
Finally,$77a+44b = 77(\frac{65}{77}) + 44(\frac{65}{44}) = 65+65 = 130$.

(B) The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.
Given $m = 2$,we have $c^2 = 4a^2 - b^2$.
The eccentricity $e$ is given by $e^2 = 1 + \frac{b^2}{a^2}$.
Substituting $e^2 = \frac{5}{2}$,we get $\frac{5}{2} = 1 + \frac{b^2}{a^2}$,which implies $\frac{b^2}{a^2} = \frac{3}{2}$,so $b^2 = \frac{3a^2}{2}$.
The length of the latus rectum is $\frac{2b^2}{a} = 6\sqrt{2}$.
Substituting $b^2 = \frac{3a^2}{2}$,we get $\frac{2}{a} \times \frac{3a^2}{2} = 6\sqrt{2}$,which simplifies to $3a = 6\sqrt{2}$,so $a = 2\sqrt{2}$.
Then $a^2 = 8$ and $b^2 = \frac{3}{2} \times 8 = 12$.
Finally,$c^2 = 4a^2 - b^2 = 4(8) - 12 = 32 - 12 = 20$.

(D) The equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
Given the eccentricity $e = \frac{\sqrt{11}}{2}$,we have $e^{2} = 1 + \frac{b^{2}}{a^{2}}$.
Substituting the value of $e$,we get $\frac{11}{4} = 1 + \frac{b^{2}}{a^{2}}$,which implies $\frac{b^{2}}{a^{2}} = \frac{7}{4}$,so $b = \frac{\sqrt{7}}{2}a$.
The sum of the lengths of the transverse axis $(2a)$ and the conjugate axis $(2b)$ is given as $2a + 2b = 4(2\sqrt{2} + \sqrt{14})$.
Substituting $b = \frac{\sqrt{7}}{2}a$,we get $2a + 2(\frac{\sqrt{7}}{2}a) = 4(2\sqrt{2} + \sqrt{14})$.
$2a + \sqrt{7}a = 4(2\sqrt{2} + \sqrt{2}\sqrt{7})$.
$a(2 + \sqrt{7}) = 4\sqrt{2}(2 + \sqrt{7})$.
Thus,$a = 4\sqrt{2}$,which means $a^{2} = 16 \times 2 = 32$.
Since $b^{2} = \frac{7}{4}a^{2}$,we have $b^{2} = \frac{7}{4} \times 32 = 7 \times 8 = 56$.
Therefore,$a^{2} + b^{2} = 32 + 56 = 88$.

(B) The equation of the circle is $x^{2} + y^{2} - 2x + 2fy + 1 = 0$. The centre of the circle is $(1, -f)$.
Since the diameters pass through the centre,we substitute $(1, -f)$ into the equations of the diameters:
$2p(1) - (-f) = 1 \implies 2p + f = 1 \implies f = 1 - 2p$.
$2(1) + p(-f) = 4p \implies 2 - pf = 4p$.
Substituting $f = 1 - 2p$ into the second equation:
$2 - p(1 - 2p) = 4p \implies 2 - p + 2p^{2} = 4p \implies 2p^{2} - 5p + 2 = 0$.
Solving for $p$: $(2p - 1)(p - 2) = 0$,so $p = 1/2$ or $p = 2$.
If $p = 1/2$,$f = 1 - 2(1/2) = 0$. Centre is $(1, 0)$.
If $p = 2$,$f = 1 - 2(2) = -3$. Centre is $(1, 3)$.
The hyperbola is $3x^{2} - y^{2} = 3$,which is $x^{2} - y^{2}/3 = 1$. The tangent line $y = mx + c$ has $c^{2} = a^{2}m^{2} - b^{2} = m^{2} - 3$.
Case $1$: Centre $(1, 0)$. $0 = m(1) + c \implies c = -m$. Then $c^{2} = m^{2} = m^{2} - 3$,which is impossible.
Case $2$: Centre $(1, 3)$. $3 = m(1) + c \implies c = 3 - m$. Then $c^{2} = (3 - m)^{2} = m^{2} - 3$.
$9 - 6m + m^{2} = m^{2} - 3 \implies 6m = 12 \implies m = 2$.

(C) The given equation of the hyperbola is $kx^{2}-y^{2}=6$,which can be written as $\frac{x^{2}}{6/k} - \frac{y^{2}}{6} = 1$.
Here,$a^{2} = \frac{6}{k}$ and $b^{2} = 6$.
The eccentricity $e$ is given by $e^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{6}{6/k} = 1 + k$.
Thus,$e = \sqrt{1+k}$.
The equation of the directrix for a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $x = \frac{a}{e}$.
Given the directrix is $x = 1$,we have $1 = \frac{\sqrt{6/k}}{\sqrt{1+k}}$.
Squaring both sides,$1 = \frac{6}{k(1+k)} \Rightarrow k^{2} + k - 6 = 0$.
Solving the quadratic equation $(k+3)(k-2) = 0$,we get $k = 2$ (since $k$ must be positive for a hyperbola).
The equation of the hyperbola is $2x^{2} - y^{2} = 6$.
Checking the options: For option $C$,$2(\sqrt{5})^{2} - (-2)^{2} = 2(5) - 4 = 10 - 4 = 6$. Thus,the point $(\sqrt{5}, -2)$ lies on the hyperbola.

(D) The tangent to the parabola $y^2 = 24x$ at $(\alpha, \beta)$ is given by $y\beta = 12(x + \alpha)$. The slope of this tangent is $m_1 = \frac{12}{\beta}$.
Given the line $2x + 2y = 5$,its slope is $m_2 = -1$. Since the tangent is perpendicular to this line,$m_1 \times m_2 = -1$,so $\frac{12}{\beta} \times (-1) = -1$,which gives $\beta = 12$.
Since $(\alpha, \beta)$ lies on $y^2 = 24x$,we have $12^2 = 24\alpha$,so $144 = 24\alpha$,which gives $\alpha = 6$.
The hyperbola is $\frac{x^2}{36} - \frac{y^2}{144} = 1$. The point on the hyperbola is $(\alpha + 4, \beta + 4) = (10, 16)$.
The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at $(x_0, y_0)$ is $\frac{a^2x}{x_0} + \frac{b^2y}{y_0} = a^2 + b^2$.
Here $a^2 = 36, b^2 = 144, x_0 = 10, y_0 = 16$. The normal is $\frac{36x}{10} + \frac{144y}{16} = 36 + 144$,which simplifies to $3.6x + 9y = 180$,or $2x + 5y = 100$.
Checking the options: For $(25, 10)$,$2(25) + 5(10) = 50 + 50 = 100$ (Passes).
For $(20, 12)$,$2(20) + 5(12) = 40 + 60 = 100$ (Passes).
For $(30, 8)$,$2(30) + 5(8) = 60 + 40 = 100$ (Passes).
For $(15, 13)$,$2(15) + 5(13) = 30 + 65 = 95 \neq 100$ (Does not pass).

(B) The rectangular hyperbola is given by $x^2-y^2=a^2$.
Let the coordinates of $B$ be $(a \sec \theta, a \tan \theta)$ and $C$ be $(a \sec \theta, -a \tan \theta)$.
Since $\triangle ABC$ is equilateral,$AB^2 = BC^2$.
The distance $BC = 2a \tan \theta$.
The distance $AB^2 = (a \sec \theta - (-a))^2 + (a \tan \theta - 0)^2 = a^2(\sec \theta + 1)^2 + a^2 \tan^2 \theta$.
Equating $AB^2 = BC^2$:
$a^2(\sec \theta + 1)^2 + a^2 \tan^2 \theta = (2a \tan \theta)^2 = 4a^2 \tan^2 \theta$.
$(\sec \theta + 1)^2 = 3 \tan^2 \theta = 3(\sec^2 \theta - 1) = 3(\sec \theta + 1)(\sec \theta - 1)$.
Since $\sec \theta + 1 \neq 0$,we have $\sec \theta + 1 = 3(\sec \theta - 1)$.
$\sec \theta + 1 = 3 \sec \theta - 3 \implies 2 \sec \theta = 4 \implies \sec \theta = 2$.
Thus,$\tan^2 \theta = \sec^2 \theta - 1 = 4 - 1 = 3$,so $\tan \theta = \sqrt{3}$.
The side length $BC = 2a \tan \theta = 2a \sqrt{3}$.
Given the side length is $ka$,we have $ka = 2a \sqrt{3}$,so $k = 2 \sqrt{3} \approx 2 \times 1.732 = 3.464$.
Since $2 < 3.464 \leq 4$,$k$ lies in the interval $(2, 4]$.

(A) Let the position of the cannon be $P$ and the speed of sound be $S$.
Let the time at which the sound of the cannon firing is heard at $A$ be $t$.
Therefore,the time at which it is heard at $B$ is $t+1$.
Hence,the distance $PA = \text{Speed} \times \text{Time} = St$.
Similarly,the distance $PB = S(t+1) = St + S$.
Subtracting the two distances,we get $PB - PA = (St + S) - St = S$.
Since $S$ is the speed of sound (a constant),$PB - PA = \text{constant}$.
By the definition of a hyperbola,the locus of a point such that the difference of its distances from two fixed points (foci) is constant is a hyperbola.
Thus,$A$ and $B$ are the foci of a hyperbola,and the cannon's position $P$ lies on one branch of this hyperbola.

(C) The equation of the hyperbola is $xy=1$,which can be written as $y = \frac{1}{x}$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{x^2}$.
The slope of the tangent at the point $(2, 1/2)$ is $\left(\frac{dy}{dx}\right)_{(2, 1/2)} = -\frac{1}{2^2} = -\frac{1}{4}$.
The slope of the normal at $(2, 1/2)$ is $n = -\frac{1}{(-1/4)} = 4$.
Let $m$ be the slope of the incident beam of light. The reflected ray is parallel to the $Y$-axis,so its slope is $\infty$.
According to the law of reflection,the angle between the incident ray and the normal is equal to the angle between the reflected ray and the normal. Using the formula for the angle between two lines with slopes $m_1$ and $m_2$,$\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$,we have:
$\left|\frac{4 - m}{1 + 4m}\right| = \left|\frac{\infty - 4}{1 + 4(\infty)}\right| = \left|\frac{1}{4}\right|$.
Solving $\frac{4 - m}{1 + 4m} = \frac{1}{4}$:
$16 - 4m = 1 + 4m$
$8m = 15 \Rightarrow m = \frac{15}{8}$.
Solving $\frac{4 - m}{1 + 4m} = -\frac{1}{4}$:
$16 - 4m = -1 - 4m$
$16 = -1$,which is impossible.
Thus,the slope of the incident beam is $\frac{15}{8}$.

(B) Given the vertices of the hyperbola are $(\pm 6, 0)$,we have $a = 6$.
Given eccentricity $e = \frac{\sqrt{5}}{2}$.
We know $b^2 = a^2(e^2 - 1) = 36 \left( \frac{5}{4} - 1 \right) = 36 \left( \frac{1}{4} \right) = 9$.
So,the equation of the hyperbola $H$ is $\frac{x^2}{36} - \frac{y^2}{9} = 1$.
The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(a \sec \theta, b \tan \theta)$ is $\frac{ax}{\sec \theta} + \frac{by}{\tan \theta} = a^2 + b^2$.
Substituting $a=6, b=3$,the equation is $\frac{6x}{\sec \theta} + \frac{3y}{\tan \theta} = 36 + 9 = 45$,which simplifies to $6x \cos \theta + 3y \cot \theta = 45$.
The slope of this normal is $-\frac{6 \cos \theta}{3 \cot \theta} = -2 \sin \theta$.
The normal is parallel to $\sqrt{2}x + y = 2\sqrt{2}$,which has a slope of $-\sqrt{2}$.
So,$-2 \sin \theta = -\sqrt{2} \Rightarrow \sin \theta = \frac{1}{\sqrt{2}} \Rightarrow \theta = \frac{\pi}{4}$.
Substituting $\theta = \frac{\pi}{4}$ into the normal equation: $6x \cos(\frac{\pi}{4}) + 3y \cot(\frac{\pi}{4}) = 45 \Rightarrow 6x(\frac{1}{\sqrt{2}}) + 3y(1) = 45 \Rightarrow 3\sqrt{2}x + 3y = 45 \Rightarrow \sqrt{2}x + y = 15$.
The point $P$ on the hyperbola is $(6 \sec(\frac{\pi}{4}), 3 \tan(\frac{\pi}{4})) = (6\sqrt{2}, 3)$.
The intersection of the normal with the $y$-axis $(x=0)$ is $K(0, 15)$.
The length $d$ is the distance between $P(6\sqrt{2}, 3)$ and $K(0, 15)$.
$d^2 = (6\sqrt{2} - 0)^2 + (3 - 15)^2 = (6\sqrt{2})^2 + (-12)^2 = 72 + 144 = 216$.

(A) The foci are given by $(h \pm ae, k) = (1 \pm \sqrt{2}, 0)$.
Comparing,we get the center $(h, k) = (1, 0)$ and $ae = \sqrt{2}$.
Given the eccentricity $e = \sqrt{2}$,we have $a(\sqrt{2}) = \sqrt{2}$,which implies $a = 1$.
For a hyperbola,$b^2 = a^2(e^2 - 1)$.
Substituting the values,$b^2 = 1^2((\sqrt{2})^2 - 1) = 1(2 - 1) = 1$.
Thus,$b = 1$.
The length of the latus rectum is given by $\frac{2b^2}{a} = \frac{2(1)^2}{1} = 2$.

(C) The equation of the hyperbola is $3x^2 - 4y^2 = 36$,which can be written as $\frac{x^2}{12} - \frac{y^2}{9} = 1$.
Here $a^2 = 12$ and $b^2 = 9$,so $a = 2\sqrt{3}$ and $b = 3$.
The parametric coordinates of a point $P$ on the hyperbola are $(a \sec \theta, b \tan \theta) = (2\sqrt{3} \sec \theta, 3 \tan \theta)$.
The slope of the tangent at $P$ must be equal to the slope of the line $3x + 2y = 1$,which is $m = -\frac{3}{2}$.
The slope of the tangent at $(x_0, y_0)$ is $\frac{dy}{dx} = \frac{3x_0}{4y_0} = -\frac{3}{2} \implies \frac{x_0}{2y_0} = -1 \implies x_0 = -2y_0$.
Substitute $x_0 = -2y_0$ into the hyperbola equation: $3(-2y_0)^2 - 4y_0^2 = 36 \implies 12y_0^2 - 4y_0^2 = 36 \implies 8y_0^2 = 36 \implies y_0^2 = \frac{9}{2} \implies y_0 = \pm \frac{3}{\sqrt{2}}$.
Since the line $3x + 2y = 1$ is in the first and fourth quadrants,we check the distance. For $y_0 = -\frac{3}{\sqrt{2}}$,$x_0 = -2(-\frac{3}{\sqrt{2}}) = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.
Then $\sqrt{2}(y_0 - x_0) = \sqrt{2}(-\frac{3}{\sqrt{2}} - \frac{6}{\sqrt{2}}) = -3 - 6 = -9$.

(B) The area of the rectangle $R$ is $2 \times 5 = 10$ square units.
The line segment $AB$ cuts off a triangle $OAB$ from the rectangle,where $O$ is the origin $(0,0)$.
The area of triangle $OAB$ is $\frac{1}{2} \times \alpha \times \beta = \frac{\alpha \beta}{2}$.
The line segment $AB$ divides the rectangle into two parts with areas in the ratio $4:1$. The area of the triangle $OAB$ is the smaller part,so its area must be $\frac{1}{5}$ of the total area of the rectangle.
$\frac{\text{Area}(OAB)}{\text{Area}(R)} = \frac{1}{5} \implies \frac{\alpha \beta / 2}{10} = \frac{1}{5} \implies \frac{\alpha \beta}{20} = \frac{1}{5} \implies \alpha \beta = 4$.
Let $M(h, k)$ be the mid-point of $AB$. Then $h = \frac{\alpha}{2}$ and $k = \frac{\beta}{2}$,which means $\alpha = 2h$ and $\beta = 2k$.
Substituting these into $\alpha \beta = 4$,we get $(2h)(2k) = 4 \implies 4hk = 4 \implies hk = 1$.
The locus of the mid-point $M(x, y)$ is $xy = 1$,which represents a hyperbola.