Analytical Geometry - Conics - Asymptotes of Hyperbola

Find the angle between the asymptotes of the hyperbola $\;$ $24x^2 - 8y^2 = 27$

Equation of hyperbola is $\;$ $24x^2 - 8y^2 = 27$

i.e. $\;$ $\dfrac{x^2}{27 / 24} - \dfrac{y^2}{27 / 8} = 1$

i.e. $\;$ $\dfrac{x^2}{9 / 8} - \dfrac{y^2}{27 / 8} = 1$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = \dfrac{9}{8}$ $\implies$ $a = \dfrac{3}{2 \sqrt{2}}$

and $b^2 = \dfrac{27}{8}$ $\implies$ $b = \dfrac{3 \sqrt{3}}{2 \sqrt{2}}$

The angle between the asymptotes is

$\begin{aligned} 2 \alpha & = 2 \tan^{-1} \left(\dfrac{b}{a}\right) \\\\ & = 2 \tan^{-1} \left(\dfrac{3 \sqrt{3} / 2 \sqrt{2}}{3 / 2 \sqrt{2}}\right) \\\\ & = 2 \tan^{-1} \left(\sqrt{3}\right) \end{aligned}$

i.e. $\;$ $2 \alpha = \dfrac{2 \pi}{3}$

Analytical Geometry - Conics - Asymptotes of Hyperbola

Find the equation of the hyperbola if its asymptotes are parallel to $\;$ $x + 2y - 12 = 0$ $\;$ and $\;$ $x - 2y+ 8 = 0$, $\;$ $\left(2,4\right)$ is the center of the hyperbola and it passes through $\left(2,0\right)$.

Equations of the given straight lines are

$x + 2y - 12 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x - 2y + 8 = 0$ $\;\;\; \cdots \; (2)$

Slope of equation $(1)$ is $= m_1 = \dfrac{-1}{2}$

Slope of equation $(2)$ is $= m_2 = \dfrac{1}{2}$

$\because$ $\;$ the asymptotes are parallel to the given lines [equations $(1)$ and $(2)$],

$\therefore$ $\;$ the slopes of the asymptotes are $\dfrac{1}{2}$ and $\dfrac{-1}{2}$

Let the equations of the asymptotes be

$y = \dfrac{1}{2}x + p$ $\;\;\; \cdots \; (3a)$ $\;$ and $\;$ $y = -\dfrac{1}{2}x + q$ $\;\;\; \cdots \; (3b)$

where $p$ and $q$ are the Y intercepts.

Solving equations $(3a)$ and $(3b)$ simultaneously gives the point of intersection of the asymptotes.

$\therefore$ $\;$ We have from equations $(3a)$ and $(3b)$

$\dfrac{1}{2}x + p = \dfrac{-1}{2}x + q$ $\implies$ $x = q - p$

Substituting the value of $x$ in equation $(3a)$ gives

$y = \dfrac{1}{2} \left(q - p\right) + p$ $\implies$ $y = \dfrac{1}{2} \left(p + q\right)$

The asymptotes intersect at the center.

Given: Center of hyperbola $= \left(2, 4\right)$

$\therefore$ $\;$ We have $\;$ $q - p = 2$ $\;\;\; \cdots \; (4a)$

and $\;$ $\dfrac{1}{2} \left(p + q\right) = 4$ $\implies$ $p + q = 8$ $\;\;\; \cdots \; (4b)$

Solving equations $(4a)$ and $(4b)$ simultaneously we have,

$2q = 10$ $\implies$ $q = 5$

Substituting the value of $q$ in equation $(4a)$ gives

$p = q - 2 = 3$

$\therefore$ $\;$ The equations of the asymptotes are

$y = \dfrac{1}{2}x + 3$ $\;\;$ i.e. $\;$ $x - 2y + 6 = 0$

and $\;$ $y = \dfrac{-1}{2}x + 5$ $\;\;$ i.e. $\;$ $x + 2y - 10 = 0$

$\therefore$ $\;$ The combined equations of the asymptotes is

$\left(x - 2y + 6\right) \left(x + 2y - 10\right) = 0$ $\;\;\; \cdots \; (5)$

The equation of the hyperbola differs from the combined equation of the asymptotes by a constant.

$\therefore$ $\;$ The equation of the hyperbola is of the form

$\left(x - 2y + 6\right) \left(x + 2y - 10\right) + k = 0$ $\;\;\; \cdots \; (6)$

The required hyperbola passes through the point $\left(2, 0\right)$.

$\therefore$ $\;$ We have from equation $(6)$,

$\left(2 - 0 + 6\right) \left(2 + 0 - 10\right) + k = 0$

$\implies$ $k = 64$

Substituting the value of $k$ in equation $(6)$, the equation of the required hyperbola is

$\left(x - 2y + 6\right) \left(x + 2y - 10\right) + 64 = 0$

Analytical Geometry - Conics - Asymptotes of Hyperbola

Find the equation of the asymptotes to the hyperbola $8x^2 + 10xy - 3y^2 - 2x + 4y -2 = 0$

Equation of given hyperbola is $\;$ $8x^2 + 10xy - 3y^2 - 2x + 4y - 2 = 0$ $\;\;\; \cdots \; (1)$

The combined equation of the asymptotes differs from the hyperbola by a constant only.

$\therefore$ $\;$ Let the combined equation of the asymptotes be: $\;$ $8x^2 + 10xy - 3y^2 - 2x + 4y + k = 0$

Now,

$\begin{aligned} 8x^2 + 10xy - 3y^2 & = 8x^2 + 12xy - 2xy - 3y^2 \\\\ & = 4x \left(2x + 3y\right) - y \left(2x + 3y\right) \\\\ & = \left(4x - y\right) \left(2x + 3y\right) \end{aligned}$

$\therefore$ $\;$ The separate equations of the asymptotes are

$4x - y + \ell = 0$ $\;\;\; \cdots \; (2a)$ and

$2x + 3y + m = 0$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ $\left(4x - y + \ell\right) \left(2x + 3y + m\right) = 8x^2 + 10xy - 3y^2 - 2x + 4y + k$ $\;\;\; \cdots \; (3)$

Equating the coefficient of the $x$ terms we have

$4 m + 2 \ell = -2$

i.e. $\;$ $2 m + \ell = -1$ $\;\;\; \cdots \; (4a)$

Equating the coefficient of the $y$ terms we have

$- m + 3 \ell = 4$ $\;\;\; \cdots \; (4b)$

Equating the constant term we have

$\ell m = k$ $\;\;\; \cdots \; (4c)$

Solving equations $(4a)$ and $(4b)$ simultaneously we get

$\ell = 1$ $\;$ and $m = -1$

$\therefore$ $\;$ We have from equation $(4c)$, $\;$ $k = -1$

$\therefore$ $\;$ The separate equations of the asymptotes are

$4x - y + 1 = 0$ $\;$ and $\;$ $2x + 3y - 1 = 0$

The combined equation of the asymptotes is

$8x^2 + 10xy - 3y^2 - 2x + 4y - 1 = 0$

Analytical Geometry - Conics - Hyperbola

Find the eccentricity, center, foci and vertices of the hyperbola $x^2 - 3y^2 + 6x + 6y + 18 = 0$ and sketch it.

Equation of given hyperbola is

$x^2 - 3y^2 + 6x + 6y + 18 = 0$

i.e. $\left(x^2 + 6x\right) - 3 \left(y^2 - 2y\right) = -18$

i.e. $\left(x^2 + 6x + 9\right) - 9 - 3 \left[\left(y^2 - 2y + 1\right) - 1\right] = -18$

i.e. $\left(x + 3\right)^2 - 3 \left(y - 1\right)^2 = -18 + 9 - 3$

i.e. $\left(x + 3\right)^2 - 3 \left(y - 1\right)^2 = -12$

i.e. $\dfrac{\left(y - 1\right)^2}{12 / 3} - \dfrac{\left(x + 3\right)^2}{12} = 1$

i.e. $\dfrac{\left(y - 1\right)^2}{4} - \dfrac{\left(x + 3\right)^2}{12} = 1$ $\;\;\; \cdots \; (1)$

Let $\;$ $Y = y - 1$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $X = x + 3$ $\;\;\; \cdots \; (2b)$

Then, we have from equations $(1)$, $(2a)$ and $(2b)$,

$\dfrac{Y^2}{4} - \dfrac{X^2}{12} = 1$ $\;\;\; \cdots \; (3)$

The transverse axis of the hyperbola given by equation $(3)$ is along the Y axis.

Comparing equation $(3)$ with the standard equation of hyperbola $\;$ $\dfrac{Y^2}{a^2} - \dfrac{X^2}{b^2} = 1$ $\;$ gives

$a^2 = 4 \implies a = 2$ $\;$ and $\;$ $b^2 = 12 \implies b = 2 \sqrt{3}$

Eccentricity $= e = \sqrt{1 + \dfrac{b^2}{a^2}} = \sqrt{1 + \dfrac{12}{4}} = \sqrt{4} = 2$

	Referred to X, Y	Referred to x, y $X = x + 3$; $\;$ $Y = y - 1$ i.e. $\;$ $x = X - 3$; $\;$ $y = Y + 1$
Center	$C \left(0, 0\right)$	$X = 0 \implies x = -3$ $Y = 0 \implies y = 1$ $\therefore$ $\;$ Center $C' = \left(-3, 1\right)$
Foci	$F_1 = \left(0, ae\right) = \left(0, 2 \times 2\right)$ i.e. $\;$ $F_1 = \left(0,4\right)$	$X = 0 \implies x = -3$ $Y = 4 \implies y = 5$ $\therefore$ $\;$ $F'_1 = \left(-3, 5\right)$
	$F_2 = \left(0, -ae\right) = \left(0, -2 \times 2\right)$ i.e. $\;$ $F_2 = \left(0, -4\right)$	$X = 0 \implies x = -3$ $Y = -4 \implies y = -3$ $\therefore$ $\;$ $F'_2 = \left(-3, -3\right)$
Vertices	$A_1 = \left(0, a\right) = \left(0, 2\right)$	$X = 0 \implies x = -3$ $Y = 2 \implies y = 3$ $\therefore$ $\;$ $A'_1 = \left(-3, 3\right)$
	$A_2 = \left(0, -a\right) = \left(0, -2\right)$	$X = 0 \implies x = -3$ $Y = -2 \implies y = -1$ $\therefore$ $\;$ $A'_2 = \left(-3, -1\right)$

Analytical Geometry - Conics - Hyperbola

Find the eccentricity, center, foci and vertices of the hyperbola $25x^2 - 16y^2 = 400$ and sketch it.

Equation of given hyperbola is: $\;$ $25x^2 - 16y^2 = 400$

i.e. $\;$ $\dfrac{x^2}{400 / 25} - \dfrac{y^2}{400 / 16} = 1$

i.e. $\;$ $\dfrac{x^2}{16} - \dfrac{y^2}{25} = 1$ $\;\;\; \cdots \; (1)$

The transverse axis is along the X axis.

Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 16 \implies a = 4$; $\;$ and $\;$ $b^2 = 25 \implies b = 5$

Now, eccentricity $= e = \sqrt{1 + \dfrac{b^2}{a^2}} = \sqrt{1 + \dfrac{25}{16}} = \dfrac{\sqrt{41}}{4}$

Center $= C \left(0, 0\right)$

Foci [$F_1$ and $F_2$] $= \left(\pm ae, 0\right) = \left(\pm 4 \times \dfrac{\sqrt{41}}{4}, 0\right) = \left(\pm \sqrt{41}, 0\right)$

Vertices [$A_1$ and $A_2$] $= \left(\pm a, 0\right) = \left(\pm 4, 0\right)$

Analytical Geometry - Conics - Hyperbola

Find the equations of directrices, latus rectums and length of latus rectum for the hyperbola $9x^2 - 4y^2 - 36x + 32y + 8 = 0$

Equation of given hyperbola is: $\;$ $9x^2 - 4y^2 - 36x + 32y + 8 = 0$

i.e. $\;$ $9 \left(x^2 - 4x\right) - 4 \left(y^2 - 8y\right) = -8$

i.e. $\;$ $9 \left[\left(x^2 - 4x + 4\right) - 4\right] - 4 \left[\left(y^2 - 8y + 16\right) - 16\right] = -8$

i.e. $\;$ $9 \left(x - 2\right)^2 - 4 \left(y - 4\right)^2 = 36 - 64 - 8$

i.e. $\;$ $9 \left(x - 2\right)^2 - 4 \left(y - 4\right)^2 = -36$

i.e. $\;$ $\dfrac{\left(y - 4\right)^2}{36 /4} - \dfrac{\left(x - 2\right)^2}{36 / 9} = 1$

i.e. $\;$ $\dfrac{\left(y - 4\right)^2}{9} - \dfrac{\left(x - 2\right)^2}{4} = 1$ $\;\;\; \cdots \; (1)$

Let $Y = y - 4$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $X = x - 2$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2b)$ equation $(1)$ becomes

$\dfrac{Y^2}{9} - \dfrac{X^2}{4} = 1$ $\;\;\; \cdots \; (3)$

For the hyperbola given by equation $(3)$, the transverse axis is along the Y axis.

Comparing equation $(3)$ with the standard equation of hyperbola $\;$ $\dfrac{Y^2}{a^2} - \dfrac{X^2}{b^2} = 1$ $\;$ gives

$a^2 = 9 \implies a = 3$ $\;$ and $\;$ $b^2 = 4 \implies b = 2$

Now, $e = \sqrt{1 + \dfrac{b^2}{a^2}} = \sqrt{1 + \dfrac{4}{9}} = \dfrac{\sqrt{13}}{3}$

	Referred to X, Y	Referred to x, y $X = x - 2$; $\;$ $Y = y - 4$ i.e. $\;$ $x = X + 2$; $\;$ $y = Y + 4$
Equations of directrices	$Y = \pm \dfrac{a}{e} = \pm \dfrac{3}{\sqrt{13} / 3}$ i.e. $\;$ $Y = \pm \dfrac{9}{\sqrt{13}}$	$Y = \pm \dfrac{9}{\sqrt{13}}$ $ \implies y = 4 \pm \dfrac{9}{\sqrt{13}}$
Equations of latus rectum	$Y = \pm a \cdot e = \pm 3 \times \dfrac{\sqrt{13}}{3}$ i.e. $\;$ $Y = \pm \sqrt{13}$	$Y = \pm \sqrt{13}$ $\implies$ $y = 4 \pm \sqrt{13}$
Length of latus rectum	$\dfrac{2b^2}{a} = \dfrac{2 \times 4}{3} = \dfrac{8}{3}$	$\dfrac{8}{3}$

Analytical Geometry - Conics - Hyperbola

Find the equations and length of transverse and conjugate axes of the hyperbola $\;$ $16x^2 - 9y^2 + 96x + 36y - 36 = 0$

Equation of given hyperbola is: $\;$ $16x^2 - 9y^2 + 96 x + 36 y - 36 = 0$

i.e. $\;$ $16 \left(x^2 + 6x\right) - 9 \left(y^2 - 4y\right) = 36$

i.e. $\;$ $16 \left[\left(x^2 + 6x + 9\right) - 9\right] - 9 \left[\left(y^2 - 4y + 4\right) - 4\right] = 36$

i.e. $\;$ $16 \left(x + 3\right)^2 - 9 \left(y - 2\right)^2 = 144 - 36 + 36$

i.e. $\;$ $16 \left(x + 3\right)^2 - 9 \left(y - 2\right)^2 = 144$

i.e. $\;$ $\dfrac{\left(x + 3\right)^2}{144 / 16} - \dfrac{\left(y - 2\right)^2}{144 / 9} = 1$

i.e. $\;$ $\dfrac{\left(x + 3\right)^2}{9} - \dfrac{\left(y - 2\right)^2}{16} = 1$ $\;\;\; \cdots \; (1)$

Let $X = x + 3$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $Y = y - 2$ $\;\;\; \cdots \; (2b)$

Then in view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes

$\dfrac{X^2}{9} - \dfrac{Y^2}{16} = 1$ $\;\;\; \cdots \; (3)$

The transverse axis of the hyperbola given by equation $(3)$ is along the X axis.

Comparing equation $(3)$ with the standard equation of hyperbola $\;$ $\dfrac{X^2}{a^2} - \dfrac{Y^2}{b^2} = 1$ $\;$ gives

$a^2 = 9 \implies a = 3$; $\;$ $b^2 = 16 \implies b = 4$

	Referred to X, Y	Referred to x, y $X = x + 3$; $\;$ $Y = y - 2$
Equation of transverse axis	X axis $\;$ $\left(Y = 0\right)$	$Y = 0 \implies y - 2 = 0$
Equation of conjugate axis	Y axis $\;$ $\left(X = 0\right)$	$X = 0 \implies x + 3 = 0$
Length of transverse axis	$2a = 2 \times 3 = 6$	$6$
Length of conjugate axis	$2b = 2 \times 4 = 8$	$8$

Analytical Geometry - Conics - Hyperbola

Find the equation of the hyperbola if the center is $\left(1, 4\right)$; one of the foci is $\left(6, 4\right)$ and the corresponding directrix is $x = \dfrac{9}{4}$

Given: Directrix of hyperbola is: $\;$ $x = \dfrac{9}{4}$

$\implies$ the directrix is parallel to the Y axis.

$\therefore$ $\;$ Let the equation of the required hyperbola be:

$\dfrac{\left(x - h\right)^2}{a^2} - \dfrac{\left(y - k\right)^2}{b^2} = 1$ $\;\;\; \cdots \; (1)$

Given: Center $C \left(h, k\right) = \left(1, 4\right)$ $\;\;\; \cdots \; (2a)$; Focus $F_1 = \left(6, 4\right)$ $\;\;\; \cdots \; (2b)$

$CF_1 = a \cdot e = \sqrt{\left(6 - 1\right)^2 + \left(4 - 4\right)^2}$

i.e. $\;$ $a \cdot e = 5$ $\;\;\; \cdots \; (3)$

Draw CZ perpendicular to the directrix.

Then, $Z = \left(\dfrac{9}{4}, 4\right)$

Distance between center and directrix is $= CZ = \dfrac{a}{e} = \sqrt{\left(\dfrac{9}{4} - 1\right)^2 + \left(4 - 4\right)^2}$

i.e. $\;$ $\dfrac{a}{e} = \dfrac{5}{4}$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ We have from equations $(3)$ and $(4)$,

$a \cdot e \times \dfrac{a}{e} = 5 \times \dfrac{5}{4}$

i.e. $\;$ $a^2 = \dfrac{25}{4}$ $\;\;\; \cdots \; (5a)$

and $\dfrac{a \cdot e}{a / e} = \dfrac{5}{5 / 4}$ $\implies$ $e^2 = 4$

Now, $b^2 = a^2 \times \left(e^2 - 1\right) = \dfrac{25}{4} \times \left(4 - 1\right)$

i.e. $\;$ $b^2 = \dfrac{75}{4}$ $\;\;\; \cdots \; (5b)$

$\therefore$ $\;$ In view of equations $(2a)$, $(5a)$ and $(5b)$, the equation of required hyperbola [equation $(1)$] becomes

$\dfrac{\left(x -1 \right)^2}{25 / 4} - \dfrac{\left(y - 4\right)^2}{75 / 4} = 1$

Analytical Geometry - Conics - Hyperbola

Find the equation of the hyperbola if foci are $\left(\pm 3, 5\right)$ and $e = 3$.

Given: Foci $F_1 = \left(3, 5\right)$ and $F_2 = \left(-3, 5\right)$; $\;$ $e = 3$

Distance between foci $= F_1 F_2 = \sqrt{\left(3 + 3\right)^2 + \left(5 - 5\right)^2}$

i.e. $\;$ $F_1 F_2 = 6$ $\;\;\; \cdots \; (1)$

But $F_1 F_2 = 2 a e$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ We have from equations $(1)$ and $(2)$,

$2a \times 3 = 6$ $\implies$ $a = 1$

Now, $b^2 = a^2 \left(e^2 - 1\right) = 1 \times \left(9 - 1\right) = 8$

The foci are $\left(\pm 3, 5\right)$ $\implies$ the transverse axis is parallel to the X axis.

$\therefore$ $\;$ Let the equation of the required hyperbola be:

$\dfrac{\left(x - h\right)^2}{a^2} - \dfrac{\left(y - k\right)^2}{b^2} = 1$ $\;\;\; \cdots \; (3)$

Center $C \left(h, k\right)$ is the midpoint of $F_1$ and $F_2$.

$\therefore$ $\;$ $C\left(h, k\right) = \left(\dfrac{3 - 3}{2}, \dfrac{5 + 5}{2}\right) = \left(0, 5\right)$

$\therefore$ $\;$ Equation of the required hyperbola is

$\dfrac{\left(x - 0\right)^2}{1^2} - \dfrac{\left(y - 5\right)^2}{8} = 1$

i.e. $\;$ $\dfrac{x^2}{1} - \dfrac{\left(y - 5\right)^2}{8} = 1$

Analytical Geometry - Conics - Hyperbola

Find the equation of the hyperbola if the center is $\left(1, -2\right)$; length of transverse axis is 8; $e = \dfrac{5}{4}$ and the transverse axis parallel to the X axis.

Given: Center $C \left(h, k\right) = \left(1, -2\right)$; $\;$ $e = \dfrac{5}{4}$

The transverse axis is parallel to the X axis.

$\therefore$ $\;$ Let the equation of the required hyperbola be: $\;$ $\dfrac{\left(x - h\right)^2}{a^2} - \dfrac{\left(y - k\right)^2}{b^2} = 1$

Length of transverse axis $= 2a = 8$ $\implies$ $a = 4$

Now, $b^2 = a^2 \left(e^2 - 1\right)$

i.e. $\;$ $b^2 = 16 \times \left(\dfrac{25}{16} - 1\right) = 16 \times \dfrac{9}{16} = 9$

$\therefore$ $\;$ Equation of required hyperbola is: $\;$ $\dfrac{\left(x -1\right)^2}{16} - \dfrac{\left(y + 2\right)^2}{9} = 1$

Analytical Geometry - Conics - Hyperbola

Find the equation of the hyperbola if the focus is $\left(2,3\right)$; the corresponding directrix is $x + 2y = 5$ and $e = 2$.

Given: Equation of directrix is $\;$ $x + 2y = 5$; $\;$ $e = 2$

Let $P \left(x, y\right)$ be any point on the hyperbola.

Draw $PM$ perpendicular to the directrix.

By definition, $\dfrac{FP}{PM} = e$

i.e. $\;$ $FP^2 = e^2 \cdot PM^2$

i.e. $\;$ $\left(x - 2\right)^2 + \left(y - 3\right)^2 = \left(2\right)^2 \times \left(\dfrac{x + 2y - 5}{\sqrt{1^2 + 2^2}}\right)^2$

i.e. $\;$ $x^2 - 4x + 4 + y^2 - 6y + 9 = \dfrac{4}{5} \left(x^2 + 4y^2 + 25 + 4xy - 10x - 20y\right)$

i.e. $\;$ $5x^2 - 20x + 5y^2 - 30y + 65 = 4x^2 + 16y^2 + 100 + 16xy - 40x - 80y$

i.e. $\;$ $x^2 - 16xy - 11y^2 + 20x + 50y - 35 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the required equation of hyperbola.