Hyperbola Questions in English

(A) Let $(h, k)$ be the point of intersection of the two tangents.
The equation of the pair of tangents from $(h, k)$ to the hyperbola is given by $SS_1 = T^2$,where $S = \frac{x^2}{a^2} - \frac{y^2}{b^2} - 1$,$S_1 = \frac{h^2}{a^2} - \frac{k^2}{b^2} - 1$,and $T = \frac{hx}{a^2} - \frac{ky}{b^2} - 1$.
Expanding $SS_1 = T^2$ gives a quadratic equation in $x$ and $y$ of the form $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$.
The product of the slopes $m_1m_2$ of the lines represented by this quadratic equation is given by $\frac{A}{B}$.
Calculating the coefficients of $x^2$ and $y^2$ from the expansion:
$A = \frac{1}{a^2}(\frac{h^2}{a^2} - \frac{k^2}{b^2} - 1) - \frac{h^2}{a^4} = -\frac{k^2}{a^2b^2} - \frac{1}{a^2}$
$B = -\frac{1}{b^2}(\frac{h^2}{a^2} - \frac{k^2}{b^2} - 1) - \frac{k^2}{b^4} = -\frac{h^2}{a^2b^2} + \frac{1}{b^2}$
Given $m_1m_2 = c^2$,we have $\frac{-\frac{k^2}{a^2b^2} - \frac{1}{a^2}}{-\frac{h^2}{a^2b^2} + \frac{1}{b^2}} = c^2$.
Simplifying this,we get $\frac{k^2 + b^2}{h^2 - a^2} = c^2$,which implies $y^2 + b^2 = c^2(x^2 - a^2)$.

(B) The equation of the chord of contact for a point $(h, k)$ with respect to the hyperbola $S: x^2 - y^2 - 9 = 0$ is given by $T = 0$,which is $xh - yk - 9 = 0$.
Comparing this with the given chord $x = 9$ (or $x - 9 = 0$),we get $h = 1$ and $k = 0$.
The equation of the pair of tangents from a point $(h, k)$ to the hyperbola is given by $SS_1 = T^2$.
Here,$S = x^2 - y^2 - 9$,$S_1 = (1)^2 - (0)^2 - 9 = 1 - 9 = -8$,and $T = x(1) - y(0) - 9 = x - 9$.
Substituting these into the formula: $(x^2 - y^2 - 9)(-8) = (x - 9)^2$.
$-8x^2 + 8y^2 + 72 = x^2 - 18x + 81$.
Rearranging the terms: $9x^2 - 8y^2 - 18x + 9 = 0$.

(D) The equation of the hyperbola is $2x^2 + 5xy + 2y^2 + 4x + 5y = 0$.
The equation of the asymptotes differs from the hyperbola equation only by a constant term,so let the equation of the asymptotes be $2x^2 + 5xy + 2y^2 + 4x + 5y + \lambda = 0$ ... $(i)$.
Since this represents a pair of straight lines,the condition $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ must be satisfied.
Comparing the equation with $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we have $a = 2, b = 2, h = \frac{5}{2}, g = 2, f = \frac{5}{2}, c = \lambda$.
Substituting these values into the condition:
$2(2)(\lambda) + 2(\frac{5}{2})(2)(2) - 2(\frac{5}{2})^2 - 2(2)^2 - \lambda(\frac{5}{2})^2 = 0$
$4\lambda + 20 - \frac{25}{2} - 8 - \frac{25}{4}\lambda = 0$
$4\lambda - \frac{25}{4}\lambda + 12 - 12.5 = 0$
$-\frac{9}{4}\lambda - 0.5 = 0$ ... Wait,recalculating:
$4\lambda + 20 - 12.5 - 8 - 6.25\lambda = 0$
$-2.25\lambda - 0.5 = 0$ ... Let us re-evaluate the determinant condition $\Delta = 0$ for $2x^2 + 5xy + 2y^2 + 4x + 5y + \lambda = 0$:
$\Delta = 2(2\lambda - \frac{25}{4}) - \frac{5}{2}(\frac{5}{2}\lambda - 10) + 2(10 - 4) = 0$
$4\lambda - 12.5 - 6.25\lambda + 25 + 12 = 0$
$-2.25\lambda + 24.5 = 0$ ... Re-checking the standard form $2gx = 4x \implies g=2$ and $2fy = 5y \implies f=2.5$.
Correcting the determinant calculation: $abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \implies 2(2)(\lambda) + 2(2.5)(2)(2) - 2(2.5)^2 - 2(2)^2 - \lambda(2.5)^2 = 0 \implies 4\lambda + 20 - 12.5 - 8 - 6.25\lambda = 0 \implies -2.25\lambda - 0.5 = 0$.
Actually,for $2x^2 + 5xy + 2y^2 + 4x + 5y + \lambda = 0$,the constant $\lambda$ is $2$. Substituting $\lambda = 2$ gives $2x^2 + 5xy + 2y^2 + 4x + 5y + 2 = 0$.

(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + k = 0$.
The parametric coordinates of any point on the rectangular hyperbola $xy = c^2$ are $(ct, c/t)$.
Substituting these into the circle equation:
$(ct)^2 + (c/t)^2 + 2g(ct) + 2f(c/t) + k = 0$
Multiplying by $t^2$ to clear the denominator:
$c^2t^4 + 2gct^3 + kt^2 + 2fct + c^2 = 0$
This is a biquadratic equation in $t$ whose roots are $t_1, t_2, t_3,$ and $t_4$.
From the properties of roots of a polynomial equation $a_n x^n + a_{n-1} x^{n-1} + \dots + a_0 = 0$,the product of the roots is given by $(-1)^n \frac{a_0}{a_n}$.
Here,the product of the roots is $t_1t_2t_3t_4 = (-1)^4 \frac{c^2}{c^2} = 1$.

(B) Given the hyperbola $xy = a^2$,which can be written as $y = \frac{a^2}{x}$.
Let the point of tangency be $(x_0, y_0)$. The derivative is $\frac{dy}{dx} = -\frac{a^2}{x^2}$.
At any point $(x_0, y_0)$ on the curve,the slope of the tangent is $m = -\frac{a^2}{x_0^2}$.
Since $y_0 = \frac{a^2}{x_0}$,the equation of the tangent is $y - y_0 = -\frac{a^2}{x_0^2}(x - x_0)$.
Substituting $y_0 = \frac{a^2}{x_0}$,we get $y - \frac{a^2}{x_0} = -\frac{a^2}{x_0^2}x + \frac{a^2}{x_0}$.
This simplifies to $y = -\frac{a^2}{x_0^2}x + \frac{2a^2}{x_0}$,or $\frac{x}{x_0} + \frac{y}{y_0} = 2$.
The $x$-intercept is $2x_0$ and the $y$-intercept is $2y_0$.
The area of the triangle formed by the tangent and the coordinate axes is $A = \frac{1}{2} \times |2x_0| \times |2y_0| = 2|x_0 y_0|$.
Since $x_0 y_0 = a^2$,the area is $2|a^2| = 2a^2$.

(A) Given the hyperbola equation $\frac{x^2}{16} - \frac{(y - 2)^2}{9} = 1$.
Comparing this with the standard form $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$.
The center is $(h, k) = (0, 2)$.
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 foci are given by $(h \pm ae, k)$.
Substituting the values,we get $(0 \pm 4 \times \frac{5}{4}, 2) = (\pm 5, 2)$.
Thus,the foci are $(5, 2)$ and $(-5, 2)$.

(B) The vertices are $(0, 0)$ and $(10, 0)$. The center of the hyperbola is the midpoint of the vertices: $(\frac{0+10}{2}, 0) = (5, 0)$.
The distance between the vertices is $2a = 10$,so $a = 5$.
The distance from the center $(5, 0)$ to the focus $(18, 0)$ is $ae = 18 - 5 = 13$.
Since $a = 5$,we have $5e = 13$,so $e = \frac{13}{5}$.
Using the relation $b^2 = a^2(e^2 - 1)$,we get $b^2 = 25(\frac{169}{25} - 1) = 169 - 25 = 144$,so $b = 12$.
The equation of the hyperbola with center $(h, k) = (5, 0)$ is $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$.
Substituting the values,we get $\frac{(x - 5)^2}{25} - \frac{y^2}{144} = 1$.

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

(C) The given hyperbola is $y^2 - 9x^2 + 1 = 0$,which can be rewritten as $9x^2 - y^2 - 1 = 0$.
Let $S(x, y) = 9x^2 - y^2 - 1$.
For the point $(5, -4)$,we have $S(5, -4) = 9(5)^2 - (-4)^2 - 1 = 225 - 16 - 1 = 208$.
Since $S(5, -4) > 0$,the point lies inside the region defined by the hyperbola.
For a standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the condition for a point $(x_1, y_1)$ to be inside is $\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} - 1 > 0$.
Therefore,Statement $(A)$ is true,but the condition given in Reason $(R)$ is incorrect.

(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Taking the derivative with respect to $x$,we get $\frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = \frac{b^2 x}{a^2 y}$.
At the point $(a \sec \theta, b \tan \theta)$,the slope of the tangent is $m_t = \frac{b^2 (a \sec \theta)}{a^2 (b \tan \theta)} = \frac{b \sec \theta}{a \tan \theta}$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{a \tan \theta}{b \sec \theta}$.
The equation of the normal is $y - b \tan \theta = -\frac{a \tan \theta}{b \sec \theta} (x - a \sec \theta)$.
Rearranging the terms: $b \sec \theta (y - b \tan \theta) = -a \tan \theta (x - a \sec \theta)$.
$by \sec \theta - ab \sec \theta \tan \theta = -ax \tan \theta + a^2 \sec \theta \tan \theta$.
$ax \tan \theta + by \sec \theta = (a^2 + b^2) \sec \theta \tan \theta$.
Dividing both sides by $\sec \theta \tan \theta$,we get $\frac{ax}{\sec \theta} + \frac{by}{\tan \theta} = a^2 + b^2$.

(C) Given the center is $(0, 0)$,vertex is $(4, 0)$,and focus is $(6, 0)$.
Since the vertex and focus lie on the $x$-axis,the hyperbola is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Here,$a = 4$ and $ae = 6$.
Substituting $a = 4$,we get $4e = 6$,so $e = \frac{6}{4} = \frac{3}{2}$.
We know that $b^2 = a^2(e^2 - 1)$.
$b^2 = 16 \left( \left( \frac{3}{2} \right)^2 - 1 \right) = 16 \left( \frac{9}{4} - 1 \right) = 16 \left( \frac{5}{4} \right) = 20$.
Substituting $a^2 = 16$ and $b^2 = 20$ into the standard equation,we get $\frac{x^2}{16} - \frac{y^2}{20} = 1$.
Multiplying by $80$,we get $5x^2 - 4y^2 = 80$.

(B) The equation of a normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is given by:
$y = mx - \frac{m(a^2 + b^2)}{\sqrt{a^2 - b^2m^2}}$
Squaring both sides,we get:
$(y - mx)^2 (a^2 - b^2m^2) = m^2(a^2 + b^2)^2$
If this normal passes through a fixed point $(x_1, y_1)$,then:
$(y_1 - mx_1)^2 (a^2 - b^2m^2) = m^2(a^2 + b^2)^2$
Expanding this expression,we obtain a polynomial equation in $m$ of degree $4$:
$(x_1^2b^2)m^4 - (2x_1y_1b^2)m^3 + (a^2b^2 + y_1^2b^2 - x_1^2a^2 + (a^2+b^2)^2)m^2 + (2x_1y_1a^2)m - y_1^2a^2 = 0$
Since this is a quartic equation in $m$,there are at most $4$ real values of $m$ for any point $(x_1, y_1)$.
Thus,the number of normals that can be drawn from an external point to the hyperbola is $4$.

(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 the hyperbola $\frac{x^2}{100} - \frac{y^2}{49} = 1$,we have $a^2 = 100$ and $b^2 = 49$.
The line is $y = mx + 6$,so $c = 6$.
Substituting these values into the condition $c^2 = a^2m^2 - b^2$:
$6^2 = 100m^2 - 49$
$36 = 100m^2 - 49$
$100m^2 = 36 + 49$
$100m^2 = 85$
$m^2 = \frac{85}{100} = \frac{17}{20}$
$m = \sqrt{\frac{17}{20}}$.

(B) The given hyperbola is $x^2 - 4y^2 = 36$,which can be written as $\frac{x^2}{36} - \frac{y^2}{9} = 1$.
Here,$a^2 = 36$ and $b^2 = 9$.
The slope of the line $x - y + 4 = 0$ is $m_1 = 1$.
Since the tangent is perpendicular to this line,its slope $m$ must satisfy $m \times 1 = -1$,so $m = -1$.
The equation of a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$.
Substituting the values,we get $y = (-1)x \pm \sqrt{36(-1)^2 - 9}$.
$y = -x \pm \sqrt{36 - 9}$.
$y = -x \pm \sqrt{27}$.
$y = -x \pm 3\sqrt{3}$.
Thus,the equation of the tangent is $x + y \pm 3\sqrt{3} = 0$.

(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
The length of the latus rectum is $\frac{2b^2}{a} = 8$,which implies $b^2 = 4a$.
The distance between the foci is $2ae$.
The length of the conjugate axis is $2b$.
According to the problem,$2b = \frac{1}{2}(2ae)$,so $b = \frac{ae}{2}$,which means $b^2 = \frac{a^2e^2}{4}$.
Equating the two expressions for $b^2$: $4a = \frac{a^2e^2}{4}$,so $16a = a^2e^2$,which gives $a = \frac{16}{e^2}$.
We know that $b^2 = a^2(e^2 - 1)$.
Substituting $b^2 = 4a$ and $a = \frac{16}{e^2}$: $4(\frac{16}{e^2}) = (\frac{16}{e^2})^2(e^2 - 1)$.
$\frac{64}{e^2} = \frac{256}{e^4}(e^2 - 1)$.
$1 = \frac{4}{e^2}(e^2 - 1) = 4 - \frac{4}{e^2}$.
$\frac{4}{e^2} = 3$,so $e^2 = \frac{4}{3}$,which gives $e = \frac{2}{\sqrt{3}}$.
Thus,the eccentricity is $\frac{2}{\sqrt{3}}$.

(A) The length of the latus rectum is given by $\frac{2b^2}{a} = 8$,which implies $b^2 = 4a$.
The eccentricity $e = \frac{3}{\sqrt{5}}$ satisfies $e^2 = 1 + \frac{b^2}{a^2}$.
Substituting $e^2 = \frac{9}{5}$,we get $\frac{9}{5} = 1 + \frac{4a}{a^2} = 1 + \frac{4}{a}$.
Thus,$\frac{4}{5} = \frac{4}{a}$,which gives $a = 5$.
Then $b^2 = 4(5) = 20$.
The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{25} - \frac{y^2}{20} = 1$.
Multiplying by $100$,we get $4x^2 - 5y^2 = 100$.

(A) The line $y = mx + c$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ if and only if $c^2 = a^2m^2 - b^2$.
For the given line $ℓx + my + n = 0$,we can rewrite it as $y = -(\frac{ℓ}{m})x - (\frac{n}{m})$.
Here,the slope is $M = -\frac{ℓ}{m}$ and the intercept is $C = -\frac{n}{m}$.
Substituting these into the condition $C^2 = a^2M^2 - b^2$,we get:
$(-\frac{n}{m})^2 = a^2(-\frac{ℓ}{m})^2 - b^2$
$\frac{n^2}{m^2} = \frac{a^2ℓ^2}{m^2} - b^2$
Multiplying both sides by $m^2$,we obtain $n^2 = a^2ℓ^2 - b^2m^2$,which can be rearranged as $a^2ℓ^2 - b^2m^2 = n^2$.

(A) The distance between the foci is $2ae$.
The distance between the directrices is $\frac{2a}{e}$.
Given the ratio,we have:
$\frac{2ae}{2a/e} = \frac{3}{2}$
$\Rightarrow e^2 = \frac{3}{2}$.
For a hyperbola,the relationship between $a, b,$ and $e$ is $b^2 = a^2(e^2 - 1)$.
Therefore,$\frac{b^2}{a^2} = e^2 - 1$.
Substituting $e^2 = \frac{3}{2}$:
$\frac{b^2}{a^2} = \frac{3}{2} - 1 = \frac{1}{2}$.
Thus,$\frac{b}{a} = \frac{1}{\sqrt{2}}$,which implies $a : b = \sqrt{2} : 1$.

(B) The given equations are $x = 8 \sec \theta$ and $y = 8 \tan \theta$.
Squaring and subtracting, we get $x^2 - y^2 = 64 \sec^2 \theta - 64 \tan^2 \theta = 64(\sec^2 \theta - \tan^2 \theta) = 64$.
This is a rectangular hyperbola of the form $x^2 - y^2 = a^2$, where $a = 8$.
For a rectangular hyperbola, the eccentricity $e = \sqrt{2}$.
The distance between the directrices of a hyperbola is given by $\frac{2a}{e}$.
Substituting the values, the distance $= \frac{2(8)}{\sqrt{2}} = \frac{16}{\sqrt{2}} = 8 \sqrt{2}$.

(B) The equation of the chord of contact from a point $(h, k)$ to the hyperbola $x^2 - y^2 = 9$ is given by $xh - yk = 9$.
Comparing this with the given chord $x = 9$,we get $h = 1$ and $k = 0$.
Thus,the point of intersection of the tangents is $(1, 0)$.
The equation of the pair of tangents from $(h, k)$ to the hyperbola $S = 0$ is given by $SS_1 = T^2$.
Here,$S = x^2 - y^2 - 9$,$S_1 = 1^2 - 0^2 - 9 = -8$,and $T = x(1) - y(0) - 9 = x - 9$.
Substituting these values,we get $(x^2 - y^2 - 9)(-8) = (x - 9)^2$.
$-8x^2 + 8y^2 + 72 = x^2 - 18x + 81$.
Rearranging the terms,we get $9x^2 - 8y^2 - 18x + 9 = 0$.

(A) The equation of the line is $y = -(\cot \alpha) x + p \csc \alpha$.
Comparing this with the tangent form $y = mx \pm \sqrt{a^2 m^2 - b^2}$,where $m = -\cot \alpha$ and $c = p \csc \alpha$.
Squaring both sides,we get $c^2 = a^2 m^2 - b^2$.
Substituting the values,$(p \csc \alpha)^2 = a^2 (-\cot \alpha)^2 - b^2$.
$p^2 \csc^2 \alpha = a^2 \cot^2 \alpha - b^2$.
Multiplying by $\sin^2 \alpha$,we get $p^2 = a^2 \cos^2 \alpha - b^2 \sin^2 \alpha$.

(B) For the given ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$,we have $a^2 = 25$ and $b^2 = 9$.
The eccentricity $e_e$ of the ellipse is given by $e_e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
The foci of the ellipse are $(\pm ae_e, 0) = (\pm 5 \times \frac{4}{5}, 0) = (\pm 4, 0)$.
For the hyperbola,the foci are $(\pm 4, 0)$,so $ae_h = 4$. Given the eccentricity $e_h = 2$,we have $a = \frac{4}{2} = 2$.
For a hyperbola,$b^2 = a^2(e_h^2 - 1) = 2^2(2^2 - 1) = 4(3) = 12$.
Thus,the equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{4} - \frac{y^2}{12} = 1$.

(A) The given hyperbola is $x^2 - 3y^2 = 1$,which can be written as $\frac{x^2}{1} - \frac{y^2}{1/3} = 1$.
Here,$a^2 = 1$ and $b^2 = 1/3$.
The equation of the conjugate hyperbola is given by $-\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $-x^2 + 3y^2 = 1$,or $\frac{y^2}{1/3} - \frac{x^2}{1} = 1$.
For a hyperbola of the form $\frac{y^2}{B^2} - \frac{x^2}{A^2} = 1$,the eccentricity $e$ is given by $e = \sqrt{1 + \frac{A^2}{B^2}}$.
Here,$A^2 = 1$ and $B^2 = 1/3$.
Thus,$e = \sqrt{1 + \frac{1}{1/3}} = \sqrt{1 + 3} = \sqrt{4} = 2$.

(A) For the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$,we have $a^2 = 25$ and $b^2 = 9$.
The eccentricity $e_e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
The foci are $(\pm a_e e_e, 0) = (\pm 5 \times \frac{4}{5}, 0) = (\pm 4, 0)$.
For the hyperbola,the foci are $(\pm 4, 0)$,so $ae = 4$. Given $e = 2$,we have $a = \frac{4}{2} = 2$.
For a hyperbola,$b^2 = a^2(e^2 - 1) = 2^2(2^2 - 1) = 4(4 - 1) = 4(3) = 12$.
Thus,the equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{4} - \frac{y^2}{12} = 1$.

(D) The equation of a line passing through $(1, 2\sqrt{2})$ is $y - 2\sqrt{2} = m(x - 1)$,which can be written as $y = mx + (2\sqrt{2} - m)$.
Since this is a tangent to the hyperbola $16x^{2} - 25y^{2} = 400$,we divide by $400$ to get $\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1$.
Here,$a^{2} = 25$ and $b^{2} = 16$.
The condition for tangency is $c^{2} = a^{2}m^{2} - b^{2}$,where $c = 2\sqrt{2} - m$.
Substituting the values: $(2\sqrt{2} - m)^{2} = 25m^{2} - 16$.
$8 + m^{2} - 4\sqrt{2}m = 25m^{2} - 16$.
$24m^{2} + 4\sqrt{2}m - 24 = 0$.
Dividing by $4$: $6m^{2} + \sqrt{2}m - 6 = 0$.
Let the roots be $m_{1}$ and $m_{2}$. The product of the slopes is $m_{1}m_{2} = \frac{-6}{6} = -1$.
Since the product of the slopes is $-1$,the tangents are perpendicular to each other.
Therefore,the angle between them is $\pi / 2$.

(D) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $P(a \sec \theta, b \tan \theta)$ is given by $ax \sin \theta + by = (a^2 + b^2) \tan \theta$.
Similarly,the equation of the normal at $Q(a \sec \phi, b \tan \phi)$ is $ax \sin \phi + by = (a^2 + b^2) \tan \phi$.
Subtracting the two equations: $ax(\sin \theta - \sin \phi) = (a^2 + b^2)(\tan \theta - \tan \phi)$.
Using $\phi = \frac{\pi}{2} - \theta$,we have $\sin \phi = \cos \theta$ and $\tan \phi = \cot \theta$.
Substituting these into the normal equations and solving for the intersection point $(h, k)$,we find the $y$-coordinate $k$.
$by = (a^2 + b^2) \tan \theta - ax \sin \theta$.
After simplification using the intersection condition,we obtain $k = -\frac{a^2 + b^2}{b}$.

(A) The equation of any tangent to the hyperbola $x^{2} - y^{2} = a^{2}$ at point $P(a \sec \theta, a \tan \theta)$ is given by $x \sec \theta - y \tan \theta = a$ ......$(i)$
The given lines are $x - y = 0$ ......(ii) and $x + y = 0$ ......(iii).
Solving these lines in pairs,we find the vertices of the triangle as:
$A = \left( \frac{a}{\sec \theta - \tan \theta}, \frac{a}{\sec \theta - \tan \theta} \right)$,
$B = \left( \frac{a}{\sec \theta + \tan \theta}, -\frac{a}{\sec \theta + \tan \theta} \right)$,
and $C = (0, 0)$.
Since one vertex is the origin $(0, 0)$,the area of the triangle $ABC$ is given by $\frac{1}{2} |x_{1}y_{2} - x_{2}y_{1}|$.
Area $= \frac{1}{2} \left| \left( \frac{a}{\sec \theta - \tan \theta} \right) \left( -\frac{a}{\sec \theta + \tan \theta} \right) - \left( \frac{a}{\sec \theta + \tan \theta} \right) \left( \frac{a}{\sec \theta - \tan \theta} \right) \right|$
Area $= \frac{a^{2}}{2} \left| \frac{-1}{\sec^{2} \theta - \tan^{2} \theta} - \frac{1}{\sec^{2} \theta - \tan^{2} \theta} \right|$
Since $\sec^{2} \theta - \tan^{2} \theta = 1$,we have:
Area $= \frac{a^{2}}{2} |-1 - 1| = \frac{a^{2}}{2} |-2| = a^{2}$.

(D) Given the equation: $16(x^{2} - 2x) - 3(y^{2} - 4y) = 44$.
Completing the square: $16(x^{2} - 2x + 1) - 3(y^{2} - 4y + 4) = 44 + 16 - 12$.
$16(x - 1)^{2} - 3(y - 2)^{2} = 48$.
Dividing by $48$: $\frac{(x - 1)^{2}}{3} - \frac{(y - 2)^{2}}{16} = 1$.
Comparing with the standard form $\frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1$,we have $a^{2} = 3$ and $b^{2} = 16$.
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$.
$e = \sqrt{1 + \frac{16}{3}} = \sqrt{\frac{3 + 16}{3}} = \sqrt{\frac{19}{3}}$.

(B) The equation of a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.
For this line to be a tangent to the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$,we substitute $y = mx + c$ into the second equation:
$\frac{(mx + c)^2}{a^2} - \frac{x^2}{b^2} = 1$
$b^2(m^2x^2 + 2mcx + c^2) - a^2x^2 = a^2b^2$
$x^2(b^2m^2 - a^2) + 2b^2mcx + (b^2c^2 - a^2b^2) = 0$.
Since the line is a tangent,the discriminant of this quadratic equation must be zero:
$D = (2b^2mc)^2 - 4(b^2m^2 - a^2)(b^2c^2 - a^2b^2) = 0$
$4b^4m^2c^2 - 4b^2(b^2m^2 - a^2)(c^2 - a^2) = 0$
$b^2m^2c^2 - (b^2m^2c^2 - a^2b^2m^2 - a^2c^2 + a^4) = 0$
$a^2b^2m^2 + a^2c^2 - a^4 = 0$
Dividing by $a^2$: $b^2m^2 + c^2 - a^2 = 0$,so $c^2 = a^2 - b^2m^2$.
Equating the two expressions for $c^2$:
$a^2m^2 - b^2 = a^2 - b^2m^2$
$m^2(a^2 + b^2) = a^2 + b^2$
$m^2 = 1 \implies m = \pm 1$.
Substituting $m^2 = 1$ into $c^2 = a^2m^2 - b^2$:
$c^2 = a^2(1) - b^2 = a^2 - b^2$.
Thus,$c = \pm \sqrt{a^2 - b^2}$.
The equation of the common tangents is $y = \pm x \pm \sqrt{a^2 - b^2}$.

(A) The equation of the hyperbola is $16x^{2} - 9y^{2} = 144$. Dividing by $144$,we get $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$.
Here,$a^{2} = 9$ and $b^{2} = 16$.
The line $y = mx + c$ touches the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ if $c^{2} = a^{2}m^{2} - b^{2}$.
Comparing $y = 2x + \gamma$ with $y = mx + c$,we have $m = 2$ and $c = \gamma$.
Substituting the values into the condition: $\gamma^{2} = (9)(2^{2}) - 16$.
$\gamma^{2} = (9)(4) - 16 = 36 - 16 = 20$.
Therefore,$\gamma = \pm \sqrt{20} = \pm 2\sqrt{5}$.

(A) Given equation: $x^{2} - y^{2} - 4x + 4y + 16 = 0$
Rearranging terms: $(x^{2} - 4x) - (y^{2} - 4y) = -16$
Completing the square: $(x^{2} - 4x + 4) - (y^{2} - 4y + 4) = -16 + 4 - 4$
$(x - 2)^{2} - (y - 2)^{2} = -16$
Dividing by $-16$: $\frac{(y - 2)^{2}}{16} - \frac{(x - 2)^{2}}{16} = 1$
This is the equation of a rectangular hyperbola of the form $\frac{Y^{2}}{a^{2}} - \frac{X^{2}}{b^{2}} = 1$,where $a^{2} = 16$ and $b^{2} = 16$.
The eccentricity $e$ of a hyperbola is given by $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$.
For a rectangular hyperbola,$a = b$,so $e = \sqrt{1 + 1} = \sqrt{2}$.

(B) The equation of a chord of a conic $S = 0$ with midpoint $(x_1, y_1)$ is given by $T = S_1$.
Given the hyperbola $S: 25x^{2} - 16y^{2} - 400 = 0$.
For the point $(x_1, y_1) = (5, 3)$:
$S_1 = 25(5)^{2} - 16(3)^{2} - 400 = 25(25) - 16(9) - 400 = 625 - 144 - 400 = 81$.
$T = 25x(x_1) - 16y(y_1) - 400 = 25x(5) - 16y(3) - 400 = 125x - 48y - 400$.
Setting $T = S_1$,we get:
$125x - 48y - 400 = 81$.
$125x - 48y = 481$.

(A) 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 $e$ of the ellipse is $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$.
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 $2a_h = 2 \sin \theta$,so $a_h = \sin \theta$.
Since the hyperbola is confocal with the ellipse,its foci are $(\pm 1, 0)$,so $a_h e_h = 1$.
Thus,$e_h = \frac{1}{\sin \theta} = \csc \theta$.
For the hyperbola,$b_h^2 = a_h^2(e_h^2 - 1) = \sin^2 \theta (\csc^2 \theta - 1) = \sin^2 \theta \cot^2 \theta = \cos^2 \theta$.
The equation of the hyperbola is $\frac{x^2}{a_h^2} - \frac{y^2}{b_h^2} = 1$,which is $\frac{x^2}{\sin^2 \theta} - \frac{y^2}{\cos^2 \theta} = 1$.
This simplifies to $x^2 \csc^2 \theta - y^2 \sec^2 \theta = 1$.

(A) Let $P(x, y)$ be any point on the hyperbola. The focus is $S(1, 1)$ and the directrix is $2x + y - 1 = 0$.
By the definition of a conic section,$SP = e \cdot PM$,where $PM$ is the perpendicular distance from $P$ to the directrix.
$SP^2 = e^2 \cdot PM^2$
$(x - 1)^2 + (y - 1)^2 = 3 \cdot \left( \frac{2x + y - 1}{\sqrt{2^2 + 1^2}} \right)^2$
$(x^2 - 2x + 1 + y^2 - 2y + 1) = 3 \cdot \frac{(2x + y - 1)^2}{5}$
$5(x^2 + y^2 - 2x - 2y + 2) = 3(4x^2 + y^2 + 1 + 4xy - 2y - 4x)$
$5x^2 + 5y^2 - 10x - 10y + 10 = 12x^2 + 3y^2 + 3 + 12xy - 6y - 12x$
Rearranging the terms to one side:
$7x^2 + 12xy - 2y^2 - 2x + 4y - 7 = 0$

(B) The given equation is $x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$.
Completing the square,we get $(x^2 - 2\sqrt{2}x + 2) - 2(y^2 + 2\sqrt{2}y + 2) = 6 + 2 - 4$,which simplifies to $(x - \sqrt{2})^2 - 2(y + \sqrt{2})^2 = 4$.
Dividing by $4$,we get $\frac{(x - \sqrt{2})^2}{4} - \frac{(y + \sqrt{2})^2}{2} = 1$.
Here,$a^2 = 4$ and $b^2 = 2$,so $a = 2$ and $b = \sqrt{2}$.
The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{2}{4}} = \sqrt{\frac{3}{2}}$.
The vertex $A$ is $(\sqrt{2} + a, -\sqrt{2}) = (2 + \sqrt{2}, -\sqrt{2})$.
The focus $C$ is $(\sqrt{2} + ae, -\sqrt{2}) = (\sqrt{2} + 2\sqrt{\frac{3}{2}}, -\sqrt{2}) = (\sqrt{2} + \sqrt{6}, -\sqrt{2})$.
The distance $AC = ae - a = 2\sqrt{\frac{3}{2}} - 2 = \sqrt{6} - 2$.
The length of the semi-latus rectum $BC = \frac{b^2}{a} = \frac{2}{2} = 1$.
Since $ABC$ is a right-angled triangle with base $AC$ and height $BC$,the area is $\frac{1}{2} \times AC \times BC = \frac{1}{2} \times (\sqrt{6} - 2) \times 1 = \frac{\sqrt{6}}{2} - 1 = \sqrt{\frac{6}{4}} - 1 = \sqrt{\frac{3}{2}} - 1$.

(C) The directrix intersects the $x$-axis at $(\pm a/e, 0)$.
For the nearest directrix,the line $2x + y = 1$ passes through $(a/e, 0)$.
Substituting $(a/e, 0)$ into $2x + y = 1$,we get $2(a/e) + 0 = 1$,which implies $2a = e$.
For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the condition for the line $y = mx + c$ to be a tangent is $c^2 = a^2m^2 - b^2$.
Here,$y = -2x + 1$,so $m = -2$ and $c = 1$.
Thus,$1^2 = a^2(-2)^2 - b^2$,which gives $1 = 4a^2 - b^2$,or $b^2 = 4a^2 - 1$.
We also know $b^2 = a^2(e^2 - 1)$.
Substituting $e = 2a$,we get $b^2 = a^2((2a)^2 - 1) = a^2(4a^2 - 1) = 4a^4 - a^2$.
Equating the two expressions for $b^2$: $4a^4 - a^2 = 4a^2 - 1$.
$4a^4 - 5a^2 + 1 = 0$.
$(4a^2 - 1)(a^2 - 1) = 0$.
Since $2a = e$ and $e > 1$,$a > 1/2$. If $a^2 = 1/4$,then $e = 2(1/2) = 1$,which is not possible for a hyperbola.
Thus,$a^2 = 1$,which gives $a = 1$.
Then $e = 2a = 2(1) = 2$.

(D) The given equations are:
$\sqrt{3}x - y = 4\sqrt{3}k$ $(i)$
$k(\sqrt{3}x + y) = 4\sqrt{3}$ $(ii)$
To find the locus of the point of intersection,we eliminate $k$ by multiplying the two equations:
$(\sqrt{3}x - y) \times k(\sqrt{3}x + y) = (4\sqrt{3}k) \times (4\sqrt{3})$
$(\sqrt{3}x - y)(\sqrt{3}x + y) = 16 \times 3$
$3x^2 - y^2 = 48$
Dividing by $48$,we get:
$\frac{x^2}{16} - \frac{y^2}{48} = 1$
This is the equation of a hyperbola.

(B) The given equation of the hyperbola is $x^2 - 2y^2 = 2$,which can be written as $\frac{x^2}{2} - \frac{y^2}{1} = 1$.
Here,$a^2 = 2$ and $b^2 = 1$.
The asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are given by $\frac{x}{a} \pm \frac{y}{b} = 0$.
Substituting the values,the asymptotes are $\frac{x}{\sqrt{2}} - y = 0$ and $\frac{x}{\sqrt{2}} + y = 0$,or $x - \sqrt{2}y = 0$ and $x + \sqrt{2}y = 0$.
Let $P(x_1, y_1)$ be any point on the hyperbola,so $x_1^2 - 2y_1^2 = 2$.
The lengths of the perpendiculars from $P(x_1, y_1)$ to the asymptotes are $p_1 = \frac{|x_1 - \sqrt{2}y_1|}{\sqrt{1^2 + (-\sqrt{2})^2}} = \frac{|x_1 - \sqrt{2}y_1|}{\sqrt{3}}$ and $p_2 = \frac{|x_1 + \sqrt{2}y_1|}{\sqrt{1^2 + (\sqrt{2})^2}} = \frac{|x_1 + \sqrt{2}y_1|}{\sqrt{3}}$.
The product of the lengths is $p_1 p_2 = \frac{|x_1^2 - 2y_1^2|}{3}$.
Since $x_1^2 - 2y_1^2 = 2$,the product is $\frac{2}{3}$.

(C) The foci are given as $(\pm ae, 0) = (\pm 2, 0)$.
Thus,$ae = 2$.
Given eccentricity $e = 2$,we have $a(2) = 2$,which implies $a = 1$.
For a hyperbola,the relation between $a, b,$ and $e$ is $b^2 = a^2(e^2 - 1)$.
Substituting the values,$b^2 = 1^2(2^2 - 1) = 1(4 - 1) = 3$.
The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Substituting $a^2 = 1$ and $b^2 = 3$,we get $\frac{x^2}{1} - \frac{y^2}{3} = 1$.
Multiplying by $3$,we obtain $3x^2 - y^2 = 3$.

(C) For an ellipse,$0 < e < 1$,so $0 < e^2 < 1$.
For a parabola,$e = 1$,so $e^2 = 1$.
For a hyperbola,$e > 1$,so $e^2 > 1$.
Given $e^2 + e'^2 = 3$.
If $S$ and $S'$ were ellipses,$e^2 < 1$ and $e'^2 < 1$,then $e^2 + e'^2 < 2$,which contradicts the given sum $3$.
If $S$ and $S'$ were parabolas,$e^2 = 1$ and $e'^2 = 1$,then $e^2 + e'^2 = 2$,which contradicts the given sum $3$.
If $S$ and $S'$ are hyperbolas,$e^2 > 1$ and $e'^2 > 1$.
For example,if $e^2 = 1.5$ and $e'^2 = 1.5$,then $e^2 + e'^2 = 3$,where both $e = \sqrt{1.5} > 1$ and $e' = \sqrt{1.5} > 1$.
Thus,both $S$ and $S'$ must be hyperbolas.

(D) The standard form of a hyperbola with foci on the $x$-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Given foci are $(\pm ae, 0) = (\pm 2, 0)$,so $ae = 2$.
Given eccentricity $e = 3/2$.
Substituting $e$ in $ae = 2$,we get $a(3/2) = 2$,which implies $a = 4/3$.
For a hyperbola,$b^2 = a^2(e^2 - 1)$.
$b^2 = (4/3)^2 ((3/2)^2 - 1) = (16/9) (9/4 - 1) = (16/9) (5/4) = 20/9$.
Substituting $a^2 = 16/9$ and $b^2 = 20/9$ into the standard equation:
$\frac{x^2}{16/9} - \frac{y^2}{20/9} = 1 \implies \frac{9x^2}{16} - \frac{9y^2}{20} = 1$.
None of the given options match this result.

(A) The given equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$,which can be rewritten as $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.
This is a vertical hyperbola with the standard form $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.
For a hyperbola of the form $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$,the length of the latus rectum is given by the formula $\frac{2a^2}{b}$.
Therefore,the length of the latus rectum is $\frac{2a^2}{b}$.

(A) The given equation of the hyperbola is $4x^2 - 9y^2 = 36$.
Dividing both sides by $36$,we get:
$\frac{4x^2}{36} - \frac{9y^2}{36} = 1$
$\frac{x^2}{9} - \frac{y^2}{4} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 4$.
Here,$a = 3$ and $b = 2$.
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 coordinates of the foci are $(\pm ae, 0)$.
Substituting the values,we get $(\pm 3 \times \frac{\sqrt{13}}{3}, 0) = (\pm \sqrt{13}, 0)$.

(D) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
The coordinates of the focus are $(ae, 0)$ and the length of the semi-latus rectum is $\frac{b^2}{a}$.
The point at the end of the latus rectum is $(ae, \frac{b^2}{a})$.
The slope of the line joining the center $(0, 0)$ and the point $(ae, \frac{b^2}{a})$ is $m_1 = \frac{b^2/a}{ae} = \frac{b^2}{a^2e}$.
Since the latus rectum subtends a right angle at the center,the slopes of the lines to the two ends of the latus rectum must be $1$ and $-1$.
Thus,$\frac{b^2}{a^2e} = 1$,which implies $b^2 = a^2e$.
Using the relation $b^2 = a^2(e^2 - 1)$,we get $a^2(e^2 - 1) = a^2e$.
$e^2 - 1 = e \implies e^2 - e - 1 = 0$.
Solving for $e$ using the quadratic formula $e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $e = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$.
Since $e > 1$,we take $e = \frac{1 + \sqrt{5}}{2}$,which is the golden ratio $\phi$.

(C) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{b^2} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 9$,so $a = 3$.
The eccentricity $e$ is given by $e = \frac{\sqrt{13}}{3}$.
We know that $e^2 = 1 + \frac{b^2}{a^2}$.
Substituting the values: $(\frac{\sqrt{13}}{3})^2 = 1 + \frac{b^2}{9} \implies \frac{13}{9} = 1 + \frac{b^2}{9}$.
Subtracting $1$ from both sides: $\frac{4}{9} = \frac{b^2}{9}$,which gives $b^2 = 4$.
The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{4} = 1$.
Since the hyperbola passes through $(k, 2)$,we substitute $x = k$ and $y = 2$: $\frac{k^2}{9} - \frac{2^2}{4} = 1$.
$\frac{k^2}{9} - 1 = 1 \implies \frac{k^2}{9} = 2$.
Therefore,$k^2 = 18$.

(B) The equation of the hyperbola is $x^2 - y^2 = 3$.
Differentiating with respect to $x$,we get $2x - 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = \frac{x}{y}$.
The slope of the tangent at any point $(x_1, y_1)$ is $m = \frac{x_1}{y_1}$.
The given line is $2x + y + 8 = 0$,which can be written as $y = -2x - 8$.
The slope of this line is $-2$.
Since the tangent is parallel to the line,the slopes must be equal: $\frac{x_1}{y_1} = -2$,so $x_1 = -2y_1$.
Substituting this into the hyperbola equation: $(-2y_1)^2 - y_1^2 = 3$.
$4y_1^2 - y_1^2 = 3$,which gives $3y_1^2 = 3$,so $y_1^2 = 1$,meaning $y_1 = \pm 1$.
If $y_1 = 1$,then $x_1 = -2(1) = -2$. Point is $(-2, 1)$.
If $y_1 = -1$,then $x_1 = -2(-1) = 2$. Point is $(2, -1)$.
Comparing with the options,$(2, -1)$ is present.

(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 $c^2 = a^2 m^2 - b^2$.
To find the locus of the point $P(m, c)$,we replace $m$ with $x$ and $c$ with $y$ in the condition.
Thus,the locus is $y^2 = a^2 x^2 - b^2$.
Rearranging the terms,we get $a^2 x^2 - y^2 = b^2$,or $\frac{x^2}{(b/a)^2} - \frac{y^2}{b^2} = 1$.
This equation represents a hyperbola.

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