Hyperbola Questions in English

(B) Let the coordinates of $Q$ be $(x, y)$.
Given $A = (1, 0)$ and $B = (-1, 0)$.
The condition is $|AQ - BQ| = 1$,which implies $AQ - BQ = \pm 1$.
Substituting the distance formula: $\sqrt{(x - 1)^2 + y^2} - \sqrt{(x + 1)^2 + y^2} = \pm 1$.
Rearranging: $\sqrt{(x - 1)^2 + y^2} = \pm 1 + \sqrt{(x + 1)^2 + y^2}$.
Squaring both sides: $(x - 1)^2 + y^2 = 1 + (x + 1)^2 + y^2 \pm 2\sqrt{(x + 1)^2 + y^2}$.
$x^2 - 2x + 1 + y^2 = 1 + x^2 + 2x + 1 + y^2 \pm 2\sqrt{(x + 1)^2 + y^2}$.
$-4x - 1 = \pm 2\sqrt{(x + 1)^2 + y^2}$.
Squaring again: $(-4x - 1)^2 = 4((x + 1)^2 + y^2)$.
$16x^2 + 8x + 1 = 4(x^2 + 2x + 1 + y^2)$.
$16x^2 + 8x + 1 = 4x^2 + 8x + 4 + 4y^2$.
$12x^2 - 4y^2 = 3$.

(D) Given the equations: $x = b \sec \phi$ and $y = a \tan \phi$.
From these,we have $\frac{x}{b} = \sec \phi$ and $\frac{y}{a} = \tan \phi$.
Using the trigonometric identity $\sec^2 \phi - \tan^2 \phi = 1$,we substitute the expressions:
$\left(\frac{x}{b}\right)^2 - \left(\frac{y}{a}\right)^2 = 1$.
This simplifies to $\frac{x^2}{b^2} - \frac{y^2}{a^2} = 1$.
This equation represents a hyperbola.

(A) The definition of a hyperbola is the locus of a point $P$ such that the absolute difference of its distances from two fixed points (foci $A$ and $B$) is a constant.
Mathematically,$|PA - PB| = 2a$,where $2a$ is a constant.
Therefore,the locus of $P$ is a hyperbola.

(A) Let the point be $P(h, k)$.
Given that $|PA - PB| = 4$.
Let $A = (3, 0)$ and $B = (-3, 0)$.
$\sqrt{(h - 3)^2 + k^2} - \sqrt{(h + 3)^2 + k^2} = \pm 4$.
Squaring both sides:
$(h - 3)^2 + k^2 = 16 + (h + 3)^2 + k^2 \mp 8\sqrt{(h + 3)^2 + k^2}$.
$h^2 - 6h + 9 + k^2 = 16 + h^2 + 6h + 9 + k^2 \mp 8\sqrt{(h + 3)^2 + k^2}$.
$-12h - 16 = \mp 8\sqrt{(h + 3)^2 + k^2}$.
Dividing by $-4$: $3h + 4 = \pm 2\sqrt{(h + 3)^2 + k^2}$.
Squaring again:
$9h^2 + 24h + 16 = 4(h^2 + 6h + 9 + k^2)$.
$9h^2 + 24h + 16 = 4h^2 + 24h + 36 + 4k^2$.
$5h^2 - 4k^2 = 20$.
Dividing by $20$: $\frac{h^2}{4} - \frac{k^2}{5} = 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{x^2}{4} - \frac{y^2}{5} = 1$.

(C) The general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Comparing the given equation $3x^2 + 7xy + 2y^2 + 5x + 5y + 2 = 0$ with the general form:
$a = 3, b = 2, c = 2, h = 7/2, g = 5/2, f = 5/2$.
Now,calculate the determinant $\Delta$:
$\Delta = (3)(2)(2) + 2(5/2)(5/2)(7/2) - 3(5/2)^2 - 2(5/2)^2 - 2(7/2)^2$
$\Delta = 12 + 2(125/8) - 3(25/4) - 2(25/4) - 2(49/4)$
$\Delta = 12 + 125/4 - 75/4 - 50/4 - 98/4$
$\Delta = 12 + (125 - 75 - 50 - 98)/4 = 12 - 98/4 = 12 - 24.5 = -12.5 \neq 0$.
Since $\Delta \neq 0$,the equation does not represent a pair of straight lines. Checking the discriminant $h^2 - ab = (7/2)^2 - (3)(2) = 49/4 - 6 = 12.25 - 6 = 6.25 > 0$. Since $h^2 - ab > 0$,it represents a hyperbola.

(C) Let $P(x, y)$ be any point on the conic.
By the definition of a conic,the distance from the focus $S(1, -1)$ to $P(x, y)$ is $SP = e \cdot PM$,where $PM$ is the perpendicular distance from $P$ to the directrix $x - y + 1 = 0$.
$SP = \sqrt{(x - 1)^2 + (y + 1)^2}$
$PM = \frac{|x - y + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y + 1|}{\sqrt{2}}$
Given $e = \sqrt{2}$,we have $SP^2 = e^2 \cdot PM^2$.
$(x - 1)^2 + (y + 1)^2 = 2 \cdot \frac{(x - y + 1)^2}{2}$
$(x^2 - 2x + 1) + (y^2 + 2y + 1) = (x - y + 1)^2$
$x^2 + y^2 - 2x + 2y + 2 = x^2 + y^2 + 1 - 2xy + 2x - 2y$
Rearranging the terms:
$2xy - 4x + 4y + 1 = 0$.

(B) The equation of the hyperbola is given by $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$.
We know the trigonometric identity $\sec^2 \theta - \tan^2 \theta = 1$.
Comparing this with the hyperbola equation,we can set $\frac{x}{A} = \sec \theta$ and $\frac{y}{B} = \tan \theta$.
Therefore,the coordinates of a point on the hyperbola are $(A \sec \theta, B \tan \theta)$.

(A) For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{\frac{a^2 + b^2}{a^2}}$.
Therefore,$\frac{1}{e^2} = \frac{a^2}{a^2 + b^2}$.
For the hyperbola $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$,the eccentricity $e_1$ is given by $e_1 = \sqrt{1 + \frac{a^2}{b^2}} = \sqrt{\frac{b^2 + a^2}{b^2}}$.
Therefore,$\frac{1}{e_1^2} = \frac{b^2}{a^2 + b^2}$.
Adding these two expressions,we get $\frac{1}{e^2} + \frac{1}{e_1^2} = \frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2} = \frac{a^2 + b^2}{a^2 + b^2} = 1$.

(B) The given equation of the hyperbola is $16x^2 - 9y^2 = 144$.
Dividing by $144$,we get $\frac{x^2}{9} - \frac{y^2}{16} = 1$,which is $\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1$.
This is a standard hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a = 3$.
By the definition of a hyperbola,the absolute difference of the distances of any point $P$ on the hyperbola from the two foci $S_1$ and $S_2$ is equal to the length of the transverse axis,which is $2a$.
Therefore,$|PS_1 - PS_2| = 2a = 2(3) = 6$.

(A) The length of the latus rectum is given by $\frac{2b^2}{a} = 8$,which implies $b^2 = 4a$.
Given the eccentricity $e = \frac{3}{\sqrt{5}}$,we use the relation $e^2 = 1 + \frac{b^2}{a^2}$.
Substituting $e^2 = \frac{9}{5}$ and $b^2 = 4a$,we get $\frac{9}{5} = 1 + \frac{4a}{a^2} = 1 + \frac{4}{a}$.
This simplifies to $\frac{4}{5} = \frac{4}{a}$,so $a = 5$.
Then $b^2 = 4(5) = 20$.
The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,which becomes $\frac{x^2}{25} - \frac{y^2}{20} = 1$.
Multiplying by $100$,we get $4x^2 - 5y^2 = 100$.

(B) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Since it passes through $(3, 0)$,we have $\frac{3^2}{a^2} - \frac{0^2}{b^2} = 1$,which gives $a^2 = 9$,so $a = 3$.
Since it passes through $(3\sqrt{2}, 2)$,we have $\frac{(3\sqrt{2})^2}{a^2} - \frac{2^2}{b^2} = 1$.
Substituting $a^2 = 9$,we get $\frac{18}{9} - \frac{4}{b^2} = 1$,which simplifies to $2 - \frac{4}{b^2} = 1$.
Thus,$\frac{4}{b^2} = 1$,so $b^2 = 4$.
The eccentricity $e$ of a hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Substituting the values,$e = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}$.

(A) Given,conjugate axis length $2b = 5 \implies b = \frac{5}{2}$.
Distance between foci $2ae = 13 \implies ae = \frac{13}{2}$.
For a hyperbola,we have the relation $b^2 = a^2(e^2 - 1) = a^2e^2 - a^2$.
Substituting the values: $(\frac{5}{2})^2 = (\frac{13}{2})^2 - a^2$.
$\frac{25}{4} = \frac{169}{4} - a^2$.
$a^2 = \frac{169}{4} - \frac{25}{4} = \frac{144}{4} = 36$.
Thus,$a^2 = 36$ and $b^2 = \frac{25}{4}$.
The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
$\frac{x^2}{36} - \frac{y^2}{25/4} = 1$.
$\frac{x^2}{36} - \frac{4y^2}{25} = 1$.
Multiplying by $900$: $25x^2 - 144y^2 = 900$.

(C) The length of the transverse axis is $2a = 7$,so $a = \frac{7}{2}$ and $a^2 = \frac{49}{4}$.
The standard equation of a hyperbola with transverse axis along the $x$-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Substituting $a^2 = \frac{49}{4}$,we get $\frac{4x^2}{49} - \frac{y^2}{b^2} = 1$.
Since the hyperbola passes through $(5, -2)$,we substitute these coordinates:
$\frac{4(5)^2}{49} - \frac{(-2)^2}{b^2} = 1$
$\frac{100}{49} - \frac{4}{b^2} = 1$
$\frac{4}{b^2} = \frac{100}{49} - 1 = \frac{51}{49}$
$b^2 = \frac{4 \times 49}{51} = \frac{196}{51}$.
Substituting $b^2$ back into the equation:
$\frac{4x^2}{49} - \frac{y^2}{(196/51)} = 1$
$\frac{4}{49}x^2 - \frac{51}{196}y^2 = 1$.
Thus,option $C$ is correct.

(C) The vertices of the hyperbola are given as $(\pm 4, 0)$,which corresponds to $(\pm a, 0)$.
Thus,$a = 4$.
The foci of the hyperbola are given as $(\pm 6, 0)$,which corresponds to $(\pm ae, 0)$.
Thus,$ae = 6$.
Substituting the value of $a$,we get $4e = 6$.
Therefore,$e = \frac{6}{4} = \frac{3}{2}$.

(A) The given equation of the hyperbola is $x^2 - y^2 = 25$.
Dividing both sides by $25$,we get $\frac{x^2}{25} - \frac{y^2}{25} = 1$.
This is in the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a^2 = 25$ and $b^2 = 25$.
Thus,$a = 5$ and $b = 5$.
The eccentricity $e$ of a hyperbola is given by the formula $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Substituting the values,$e = \sqrt{1 + \frac{25}{25}} = \sqrt{1 + 1} = \sqrt{2}$.

(C) Given equation: $16x^2 - y^2 + 64x + 4y + 44 = 0$
Rearranging terms: $16(x^2 + 4x) - (y^2 - 4y) = -44$
Completing the square: $16(x^2 + 4x + 4) - (y^2 - 4y + 4) = -44 + 64 - 4$
$16(x + 2)^2 - (y - 2)^2 = 16$
Dividing by $16$: $\frac{(x + 2)^2}{1} - \frac{(y - 2)^2}{16} = 1$
This is a horizontal hyperbola with center $(-2, 2)$.
The transverse axis is parallel to the $x$-axis,given by $y = 2$.
The conjugate axis is parallel to the $y$-axis,given by $x = -2$ (or $x + 2 = 0$).

(A) The length of the transverse axis of a hyperbola is given by $2a = 8$,which implies $a = 4$.
The length of the conjugate axis is given by $2b = 6$,which implies $b = 3$.
By the definition of a hyperbola,the absolute difference of the focal distances of any point on the hyperbola is equal to the length of the transverse axis,which is $2a$.
Therefore,the difference of the focal distances is $2a = 8$.

(C) The foci are given as $(0, \pm 4)$,which corresponds to $(0, \pm be)$. Thus,$be = 4$.
The vertices are given as $(0, \pm 2)$,which corresponds to $(0, \pm b)$. Thus,$b = 2$.
Substituting $b = 2$ into $be = 4$,we get $2e = 4$,so $e = 2$.
Using the relation $a^2 = b^2(e^2 - 1)$ for a vertical hyperbola,we have $a^2 = 2^2(2^2 - 1) = 4(4 - 1) = 4(3) = 12$.
The standard equation for a vertical hyperbola is $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.
Substituting the values,we get $\frac{y^2}{4} - \frac{x^2}{12} = 1$.

(C) Given equations are:
$bxt - ayt = ab$ --- $(1)$
$bx + ay = abt$ --- $(2)$
From $(2)$,we have $t = \frac{bx + ay}{ab}$.
Substitute $t$ into $(1)$:
$bx(\frac{bx + ay}{ab}) - ay(\frac{bx + ay}{ab}) = ab$
Multiply by $ab$:
$bx(bx + ay) - ay(bx + ay) = (ab)^2$
$(bx - ay)(bx + ay) = (ab)^2$
$(bx)^2 - (ay)^2 = (ab)^2$
$b^2x^2 - a^2y^2 = a^2b^2$
Divide by $a^2b^2$:
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
This is the standard equation of a hyperbola.

(D) Given equations are:
$ax \sec \theta + by \tan \theta = a$ $(1)$
$ax \tan \theta + by \sec \theta = b$ $(2)$
Squaring both equations:
$(ax \sec \theta + by \tan \theta)^2 = a^2$
$(ax \tan \theta + by \sec \theta)^2 = b^2$
Subtracting the second from the first:
$(ax \sec \theta + by \tan \theta)^2 - (ax \tan \theta + by \sec \theta)^2 = a^2 - b^2$
$a^2 x^2 (\sec^2 \theta - \tan^2 \theta) + b^2 y^2 (\tan^2 \theta - \sec^2 \theta) = a^2 - b^2$
Since $\sec^2 \theta - \tan^2 \theta = 1$,we have:
$a^2 x^2 - b^2 y^2 = a^2 - b^2$
This is the equation of a hyperbola.

(C) Given: Centre $(h, k) = (0, 0)$,Vertex $(a, 0) = (4, 0)$,and Focus $(ae, 0) = (6, 0)$.
From the vertex,$a = 4$.
From the focus,$ae = 6$. Substituting $a = 4$,we get $4e = 6$,so $e = \frac{6}{4} = \frac{3}{2}$.
For a hyperbola,$b^2 = a^2(e^2 - 1)$.
$b^2 = 16 \left( (\frac{3}{2})^2 - 1 \right) = 16 (\frac{9}{4} - 1) = 16 (\frac{5}{4}) = 20$.
The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Substituting $a^2 = 16$ and $b^2 = 20$,we get $\frac{x^2}{16} - \frac{y^2}{20} = 1$.
Multiplying by $80$,we get $5x^2 - 4y^2 = 80$.

(B) For any hyperbola,the eccentricity $e$ must satisfy the condition $e > 1$.
We evaluate the given options:
$A) \sqrt{\frac{9}{5}} \approx 1.34 > 1$
$B) 2\sqrt{\frac{1}{9}} = 2 \times \frac{1}{3} = \frac{2}{3} \approx 0.67 < 1$
$C) 3\sqrt{\frac{1}{8}} = 3 \times \frac{1}{2\sqrt{2}} = \frac{3}{2.828} \approx 1.06 > 1$
$D) 2 > 1$
Since the eccentricity of a hyperbola cannot be less than or equal to $1$,the value $\frac{2}{3}$ is impossible. Thus,option $B$ is the correct answer.

(A) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Since it passes through $(3, 2)$:
$\frac{9}{a^2} - \frac{4}{b^2} = 1$ ..... $(i)$
Since it passes through $(-17, 12)$:
$\frac{289}{a^2} - \frac{144}{b^2} = 1$ ..... $(ii)$
Let $u = \frac{1}{a^2}$ and $v = \frac{1}{b^2}$. The equations become:
$9u - 4v = 1$ ..... $(iii)$
$289u - 144v = 1$ ..... $(iv)$
Multiply $(iii)$ by $36$:
$324u - 144v = 36$ ..... $(v)$
Subtract $(iv)$ from $(v)$:
$(324 - 289)u = 36 - 1$
$35u = 35 \implies u = 1 \implies a^2 = 1 \implies a = 1$.
Substitute $u=1$ into $(iii)$:
$9(1) - 4v = 1 \implies 4v = 8 \implies v = 2 \implies b^2 = \frac{1}{2}$.
The length of the transverse axis is $2a = 2(1) = 2$.

(C) Given lines are:
$L_1: \sqrt{3}x - y = 4\sqrt{3}k$ $(i)$
$L_2: \sqrt{3}kx + ky = 4\sqrt{3}$ (ii)
From $(i)$,$k = \frac{\sqrt{3}x - y}{4\sqrt{3}}$.
Substitute $k$ into (ii):
$\sqrt{3}x(\frac{\sqrt{3}x - y}{4\sqrt{3}}) + y(\frac{\sqrt{3}x - y}{4\sqrt{3}}) = 4\sqrt{3}$
Multiply by $4\sqrt{3}$:
$\sqrt{3}x(\sqrt{3}x - y) + y(\sqrt{3}x - y) = 48$
$3x^2 - \sqrt{3}xy + \sqrt{3}xy - y^2 = 48$
$3x^2 - y^2 = 48$
Dividing by $48$:
$\frac{x^2}{16} - \frac{y^2}{48} = 1$
This is the equation of a hyperbola.

(A) The given equation of the hyperbola is $9x^2 - 16y^2 = 144$.
Dividing both sides by $144$,we get $\frac{x^2}{16} - \frac{y^2}{9} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 16$,which implies $a = 4$.
The difference of the focal distances of any point on a hyperbola is equal to the length of its transverse axis,which is $2a$.
Therefore,the difference is $2 \times 4 = 8$.

(C) The given equation of the hyperbola is $4x^2 - 9y^2 = 16$.
Dividing by $16$,we get $\frac{x^2}{4} - \frac{y^2}{16/9} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 4$ and $b^2 = \frac{16}{9}$.
Thus,$a = 2$ and $b = \frac{4}{3}$.
The eccentricity $e$ is given by the formula $b^2 = a^2(e^2 - 1)$.
Substituting the values,$\frac{16}{9} = 4(e^2 - 1)$.
$\frac{4}{9} = e^2 - 1$.
$e^2 = 1 + \frac{4}{9} = \frac{13}{9}$.
Therefore,$e = \frac{\sqrt{13}}{3}$.

(D) The given equation of the conic is ${x^2} - 4{y^2} = 1$,which can be written in the standard form of a hyperbola as $\frac{x^2}{1^2} - \frac{y^2}{(1/2)^2} = 1$.
Here,$a^2 = 1$ and $b^2 = \frac{1}{4}$.
The eccentricity $e$ of a hyperbola is given by the formula $b^2 = a^2(e^2 - 1)$.
Substituting the values,we get $\frac{1}{4} = 1(e^2 - 1)$.
$e^2 = 1 + \frac{1}{4} = \frac{5}{4}$.
Therefore,$e = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$.

(A) The given equation is $2x^2 - 3y^2 = 5$.
Dividing by $5$,we get $\frac{x^2}{5/2} - \frac{y^2}{5/3} = 1$.
This is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a^2 = \frac{5}{2}$ and $b^2 = \frac{5}{3}$.
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{5/3}{5/2}} = \sqrt{1 + \frac{2}{3}} = \sqrt{\frac{5}{3}}$.
The foci are at $(\pm ae, 0)$.
$ae = \sqrt{\frac{5}{2}} \times \sqrt{\frac{5}{3}} = \sqrt{\frac{25}{6}} = \frac{5}{\sqrt{6}}$.
Thus,the foci are $\left( \pm \frac{5}{\sqrt{6}}, 0 \right)$.

(B) The given equation of the hyperbola is $16x^2 - 9y^2 = 144$.
Dividing both sides by $144$,we get:
$\frac{x^2}{9} - \frac{y^2}{16} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 16$.
Thus,$a = 3$ and $b = 4$.
The length of the latus-rectum is given by the formula $\frac{2b^2}{a}$.
Substituting the values,we get:
$L.R. = \frac{2 \times 16}{3} = \frac{32}{3}$.

(C) The given equation of the hyperbola is $9x^2 - 16y^2 = 144$.
Dividing both sides by $144$,we get $\frac{x^2}{16} - \frac{y^2}{9} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$.
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.
The coordinates of the foci are $(\pm ae, 0)$.
Substituting the values,we get $(\pm 4 \times \frac{5}{4}, 0) = (\pm 5, 0)$.

(A) The given equation is $3x^2 - 4y^2 = 32$.
Dividing both sides by $32$,we get $\frac{3x^2}{32} - \frac{4y^2}{32} = 1$,which simplifies to $\frac{x^2}{32/3} - \frac{y^2}{8} = 1$.
This is in the standard form of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a^2 = \frac{32}{3}$.
Thus,$a = \sqrt{\frac{32}{3}} = \frac{\sqrt{16 \times 2}}{\sqrt{3}} = \frac{4\sqrt{2}}{\sqrt{3}}$.
The length of the transverse axis is $2a$.
Therefore,$2a = 2 \times \frac{4\sqrt{2}}{\sqrt{3}} = \frac{8\sqrt{2}}{\sqrt{3}}$.

(A) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{4} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 9$ and $b^2 = 4$.
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}$.
The equation of the directrix for a hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $x = \pm \frac{a}{e}$.
Substituting the values,$x = \pm \frac{3}{\sqrt{13}/3} = \pm \frac{9}{\sqrt{13}}$.
Thus,the directrix is $x = 9/\sqrt{13}$.

(C) Given equations are:
$\frac{x}{a} - \frac{y}{b} = m$ .....$(i)$
$\frac{x}{a} + \frac{y}{b} = \frac{1}{m}$ .....$(ii)$
Multiplying equation $(i)$ and $(ii)$ gives:
$\left( \frac{x}{a} - \frac{y}{b} \right) \left( \frac{x}{a} + \frac{y}{b} \right) = m \times \frac{1}{m}$
Using the identity $(A-B)(A+B) = A^2 - B^2$,we get:
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
This is the standard equation of a hyperbola.

(D) By definition,a hyperbola is the locus of a point in a plane such that the absolute difference of its distances from two fixed points (called the foci) is a constant value equal to the length of the transverse axis $(2a)$.

(D) The given equation of the hyperbola is $2x^2 - y^2 = 6$.
Dividing both sides by $6$,we get $\frac{x^2}{3} - \frac{y^2}{6} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 3$ and $b^2 = 6$.
The eccentricity $e$ of a hyperbola is given by the formula $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Substituting the values,$e = \sqrt{1 + \frac{6}{3}} = \sqrt{1 + 2} = \sqrt{3}$.
Thus,the eccentricity is $\sqrt{3}$.

(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Given that the distance between the foci is double the distance between its vertices: $2ae = 2(2a) \implies ae = 2a \implies e = 2$.
We know that $e^2 = 1 + \frac{b^2}{a^2}$,so $4 = 1 + \frac{b^2}{a^2} \implies \frac{b^2}{a^2} = 3 \implies b^2 = 3a^2$.
The length of the conjugate axis is $2b = 6$,which implies $b = 3$,so $b^2 = 9$.
Substituting $b^2 = 9$ into $b^2 = 3a^2$,we get $9 = 3a^2 \implies a^2 = 3$.
The equation of the hyperbola is $\frac{x^2}{3} - \frac{y^2}{9} = 1$,which simplifies to $3x^2 - y^2 = 9$.

(C) The given equation is $13[(x - 1)^2 + (y - 2)^2] = 3(2x + 3y - 2)^2$.
This is of the form $SP^2 = e^2 PM^2$,where $S$ is the focus,$P$ is a point $(x, y)$,and $PM$ is the perpendicular distance from $P$ to the directrix.
Expanding the equation:
$13(x^2 - 2x + 1 + y^2 - 4y + 4) = 3(4x^2 + 9y^2 + 4 + 12xy - 8x - 12y)$
$13x^2 + 13y^2 - 26x - 52y + 65 = 12x^2 + 27y^2 + 36xy - 24x - 36y + 12$
$x^2 - 36xy - 14y^2 - 2x - 16y + 53 = 0$.
Comparing this with the general second-degree equation $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$,we have $A = 1$,$H = -18$,and $B = -14$.
The discriminant $H^2 - AB = (-18)^2 - (1)(-14) = 324 + 14 = 338$.
Since $H^2 - AB > 0$,the conic section represents a hyperbola.

(A) By the definition of a conic section,the distance from a point $P(x, y)$ to the focus $S(2, 1)$ is equal to $e$ times the perpendicular distance from $P$ to the directrix $x + 2y - 1 = 0$.
$SP^2 = e^2 \times \text{dist}(P, \text{directrix})^2$
$(x - 2)^2 + (y - 1)^2 = 2^2 \times \left[ \frac{x + 2y - 1}{\sqrt{1^2 + 2^2}} \right]^2$
$(x^2 - 4x + 4) + (y^2 - 2y + 1) = 4 \times \frac{(x + 2y - 1)^2}{5}$
$5(x^2 + y^2 - 4x - 2y + 5) = 4(x^2 + 4y^2 + 1 + 4xy - 2x - 4y)$
$5x^2 + 5y^2 - 20x - 10y + 25 = 4x^2 + 16y^2 + 4 + 16xy - 8x - 16y$
$x^2 - 16xy - 11y^2 - 12x + 6y + 21 = 0$.

(C) Given equation: $x^2 + 2x - y^2 + 5 = 0$
Completing the square for $x$: $(x^2 + 2x + 1) - y^2 + 5 - 1 = 0$
$(x + 1)^2 - y^2 = -4$
Dividing by $-4$: $\frac{y^2}{4} - \frac{(x + 1)^2}{4} = 1$
This is a hyperbola of the form $\frac{y^2}{b^2} - \frac{(x-h)^2}{a^2} = 1$,where $a^2 = 4$ and $b^2 = 4$.
Here,$a = 2$ and $b = 2$.
The eccentricity $e = \sqrt{1 + \frac{a^2}{b^2}} = \sqrt{1 + \frac{4}{4}} = \sqrt{2}$.
The equations of the directrices for this hyperbola are $y - k = \pm \frac{b}{e}$.
Since $k = 0$,$y = \pm \frac{2}{\sqrt{2}} = \pm \sqrt{2}$.

(B) The given equation of the hyperbola is $9x^2 - 16y^2 + 18x + 32y - 151 = 0$.
Rearranging the terms,we get:
$9(x^2 + 2x) - 16(y^2 - 2y) = 151$.
Completing the square for $x$ and $y$:
$9(x^2 + 2x + 1) - 16(y^2 - 2y + 1) = 151 + 9 - 16$.
$9(x + 1)^2 - 16(y - 1)^2 = 144$.
Dividing by $144$:
$\frac{(x + 1)^2}{16} - \frac{(y - 1)^2}{9} = 1$.
Comparing this with the standard form $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$,the centre $(h, k)$ is $(-1, 1)$.

(A) The foci are $(6, 4)$ and $(-4, 4)$.
Since the $y$-coordinates are the same,the transverse axis is horizontal.
The centre is the midpoint of the foci: $\left( \frac{6-4}{2}, \frac{4+4}{2} \right) = (1, 4)$.
The distance between the foci is $2ae = 6 - (-4) = 10$,so $ae = 5$.
Given $e = 2$,we have $a(2) = 5$,which gives $a = \frac{5}{2}$.
For a hyperbola,$b^2 = a^2(e^2 - 1) = \left( \frac{25}{4} \right)(4 - 1) = \frac{75}{4}$.
The equation of the hyperbola is $\frac{(x - 1)^2}{a^2} - \frac{(y - 4)^2}{b^2} = 1$.
Substituting the values: $\frac{(x - 1)^2}{25/4} - \frac{(y - 4)^2}{75/4} = 1$.
Multiplying by $75$: $3(x - 1)^2 - (y - 4)^2 = 75$.
$3(x^2 - 2x + 1) - (y^2 - 8y + 16) = 75$.
$3x^2 - 6x + 3 - y^2 + 8y - 16 = 75$.
$3x^2 - y^2 - 6x + 8y - 88 = 0$.
Wait,checking the options provided,the correct form is $12x^2 - 4y^2 - 24x + 32y - 127 = 0$.

(A) The auxiliary circle of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is defined as the circle described on the transverse axis as its diameter.
Since the length of the transverse axis is $2a$,the radius of the auxiliary circle is $a$.
The center of the hyperbola is at the origin $(0, 0)$.
Therefore,the equation of the auxiliary circle is $(x - 0)^2 + (y - 0)^2 = a^2$,which simplifies to $x^2 + y^2 = a^2$.

(C) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Comparing this with the given equation ${x^2} - 16xy - 11{y^2} - 12x + 6y + 21 = 0$,we get:
$a = 1, h = -8, b = -11, g = -6, f = 3, c = 21$.
First,we calculate the discriminant $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2$:
$\Delta = (1)(-11)(21) + 2(3)(-6)(-8) - (1)(3)^2 - (-11)(-6)^2 - (21)(-8)^2$
$\Delta = -231 + 288 - 9 + 396 - 1344 = -1000 \ne 0$.
Next,we check the condition $h^2 - ab$:
$h^2 - ab = (-8)^2 - (1)(-11) = 64 + 11 = 75$.
Since $h^2 - ab > 0$,the equation represents a hyperbola.

(D) Given equation: $9x^2 - 16y^2 - 18x - 32y - 151 = 0$
Rearranging terms: $9(x^2 - 2x) - 16(y^2 + 2y) = 151$
Completing the square: $9(x^2 - 2x + 1) - 16(y^2 + 2y + 1) = 151 + 9 - 16$
$9(x - 1)^2 - 16(y + 1)^2 = 144$
Dividing by $144$: $\frac{(x - 1)^2}{16} - \frac{(y + 1)^2}{9} = 1$
Comparing with standard form $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$,we get $a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$.
The length of the latus rectum is given by $\frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{18}{4} = \frac{9}{2}$.

(C) Given equation: $x^2 - 4y^2 - 2x + 16y - 40 = 0$
Rearranging the terms:
$(x^2 - 2x) - 4(y^2 - 4y) = 40$
Completing the square:
$(x^2 - 2x + 1) - 4(y^2 - 4y + 4) = 40 + 1 - 16$
$(x - 1)^2 - 4(y - 2)^2 = 25$
Dividing by $25$:
$\frac{(x - 1)^2}{25} - \frac{(y - 2)^2}{25/4} = 1$
This is of the form $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$,which represents a hyperbola.

(C) The given parametric equations are $x = 8 \sec \theta$ and $y = 8 \tan \theta$.
We know that $\sec^2 \theta - \tan^2 \theta = 1$.
Substituting the values,we get $\left(\frac{x}{8}\right)^2 - \left(\frac{y}{8}\right)^2 = 1$,which simplifies to $\frac{x^2}{64} - \frac{y^2}{64} = 1$.
This is a rectangular hyperbola where $a^2 = 64$ and $b^2 = 64$,so $a = 8$ and $b = 8$.
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{64}{64}} = \sqrt{2}$.
The distance between the directrices of a hyperbola is given by $\frac{2a}{e}$.
Substituting the values,we get $\frac{2 \times 8}{\sqrt{2}} = \frac{16}{\sqrt{2}} = 8 \sqrt{2}$.

(B) Given equation of the hyperbola is $5x^2 - 4y^2 + 20x + 8y = 4$.
Rearranging the terms,we get $5(x^2 + 4x) - 4(y^2 - 2y) = 4$.
Completing the square,we have $5(x^2 + 4x + 4) - 4(y^2 - 2y + 1) = 4 + 20 - 4$.
This simplifies to $5(x + 2)^2 - 4(y - 1)^2 = 20$.
Dividing by $20$,we get $\frac{(x + 2)^2}{4} - \frac{(y - 1)^2}{5} = 1$.
Comparing this with the standard form $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$,we have $a^2 = 4$ and $b^2 = 5$.
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.
$e = \sqrt{1 + \frac{5}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.

(A) The given equation of the hyperbola is $9x^2 - 16y^2 + 72x - 32y - 16 = 0$.
Rearranging the terms,we get $9(x^2 + 8x) - 16(y^2 + 2y) = 16$.
Completing the square,we have $9(x^2 + 8x + 16) - 16(y^2 + 2y + 1) = 16 + 144 - 16$.
This simplifies to $9(x + 4)^2 - 16(y + 1)^2 = 144$.
Dividing by $144$,we get $\frac{(x + 4)^2}{16} - \frac{(y + 1)^2}{9} = 1$.
Comparing this with the standard form $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$,we identify $a^2 = 16$ (so $a = 4$) and $b^2 = 9$.
The length of the latus rectum is given by $\frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{18}{4} = \frac{9}{2}$.

(C) The given hyperbola is $5x^2 - 9y^2 = 45$. Dividing by $45$,we get $\frac{x^2}{9} - \frac{y^2}{5} = 1$.
Comparing with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 5$.
The tangent line is $y = mx + c$,where $m = 1$ and $c = 2$.
The point of contact $(x_1, y_1)$ for a tangent $y = mx + c$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\left( \frac{-a^2m}{c}, \frac{-b^2}{c} \right)$.
Substituting the values: $x_1 = \frac{-9(1)}{2} = -\frac{9}{2}$ and $y_1 = \frac{-5}{2}$.
Thus,the point of contact is $(-9/2, -5/2)$.

(A) 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 the line $lx + my + n = 0$,we can rewrite it as $y = -\frac{l}{m}x - \frac{n}{m}$.
Comparing this with $y = Mx + C$,we get $M = -\frac{l}{m}$ and $C = -\frac{n}{m}$.
Substituting these values into the condition $C^2 = a^2M^2 - b^2$:
$(-\frac{n}{m})^2 = a^2(-\frac{l}{m})^2 - b^2$
$\frac{n^2}{m^2} = \frac{a^2l^2}{m^2} - b^2$
Multiplying both sides by $m^2$,we get $n^2 = a^2l^2 - b^2m^2$,which can be written as $a^2l^2 - b^2m^2 = n^2$.