Analytical Geometry - Conics - Rectangular Hyperbola

Prove that the tangent at any point to the rectangular hyperbola forms with the asymptotes a triangle of constant area.

Let the rectangular hyperbola be: $\;$ $xy = c^2$ $\;\;\; \cdots \; (1)$

The asymptotes of the rectangular hyperbola given by equation $(1)$ are the $X$ and $Y$ coordinate axes.

Any point $P \left(t\right)$ on equation $(1)$ is $\;$ $P \left(t\right) = P \left(ct, \dfrac{c}{t}\right)$

Equation of tangent at point $P \left(t\right)$ is

$x + yt^2 = 2ct$ $\;\;\; \cdots \; (2)$

Putting $y = 0$ in equation $(2)$, we get the coordinates of point $A$ as $A \left(2ct, 0\right)$.

Putting $x = 0$ in equation $(2)$, we get the coordinates of point $B$ as $B \left(0, \dfrac{2c}{t}\right)$

$\begin{aligned} \therefore \; \text{Area of } \triangle AOB & = \dfrac{1}{2} \times OA \times OB \\\\ & = \dfrac{1}{2} \times 2 ct \times \dfrac{2c}{t} \\\\ & = 2c^2 = \text{constant} \end{aligned}$

$\therefore$ $\;$ The tangent at any point to the rectangular hyperbola forms with the asymptotes a triangle of constant area.

Hence proved.

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equations of the asymptotes for the rectangular hyperbola $6x^2 + 5xy - 6y^2 + 12x + 5y + 3 = 0$

Equation of the rectangular hyperbola is

$6x^2 + 5xy - 6y^2 + 12 x + 5y + 3 = 0$ $\;\;\; \cdots \; (1)$

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

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

$6x^2 + 5xy - 6y^2 + 12x + 5y + k = 0$

Consider

$\begin{aligned} 6x^2 + 5xy - 6y^2 & = 6x^2 + 9xy - 4xy - 6y^2 \\\\ & = 3x \left(2x + 3y\right) - 2y \left(2x + 3y\right) \\\\ & = \left(3x - 2y\right) \left(2x + 3y\right) \end{aligned}$

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

$3x - 2y + \ell = 0$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $2x + 3y + m = 0$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ $\left(3x - 2y + \ell\right) \left(2x + 3y + m\right) = 6x^2 + 5xy -6y^2 + 12x + 5y + k$

Equating the coefficients of $x$ terms we get: $\;$ $3m + 2 \ell = 12$ $\;\;\; \cdots \; (3a)$

Equating the coefficients of $y$ terms we get: $\;$ $-2m + 3 \ell = 5$ $\;\;\; \cdots \; (3b)$

Equating the constant terms we get: $\;$ $\ell \times m = k$ $\;\;\; \cdots \; (3c)$

Solving equations $(3a)$ and $(3b)$ simultaneously we get,

$\ell = 3$ $\;$ and $\;$ $m = 2$

Substituting the values of $\ell$ and $m$ in equations $(2a)$ and $(2b)$ respectively, we get the separate equations of asymptotes as

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

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equations of the asymptotes for the rectangular hyperbola $2xy + 3x + 4y + 1 = 0$

Equation of given rectangular hyperbola is

$2xy + 3x + 4y + 1 = 0$

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

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

i.e. $\;$ $xy + \dfrac{3}{2}x + 2y + \dfrac{7}{2} = 3$

i.e. $\;$ $y \left(x + 2\right) + \dfrac{3}{2} \left(x + 2\right) = 3$

i.e. $\;$ $\left(x + 2\right) \left(y + \dfrac{3}{2}\right) - 3 = 0$

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

$\therefore$ $\;$ The combined equation of asymptotes is

$\left(x + 2\right) \left(y + \dfrac{3}{2}\right) = 0$

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

$x+ 2 = 0$ $\;$ and $\;$ $y + \dfrac{3}{2} = 0$

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equation of the rectangular hyperbola which has its center at $\left(2,1\right)$, one of its asymptotes $3x - y - 5 = 0$ and which passes through the point $\left(1, -1\right)$.

Let the equation of the required rectangular hyperbola be

$\left(x - h\right) \left(y - k\right) = c^2$ $\;\;\; \cdots \; (1)$

One of the asymptotes of the required rectangular hyperbola is

$3x - y - 5 = 0$ $\;\;\; \cdots \; (2)$

Slope of the given asymptote is $= m_1 = 3$

The asymptotes of a rectangular hyperbola are perpendicular to each other.

$\therefore$ $\;$ Slope of the second asymptote $= m_2 = \dfrac{-1}{m_1} = \dfrac{-1}{3}$

Let the equation of the second asymptote be

$y = \dfrac{-1}{3}x + p$ $\;$ where $p$ is a constant.

i.e. $\;$ $x + 3y = 3p$ $\;\;\; \cdots \; (3)$

Solving equations $(2)$ and $(3)$ simultaneously we have

$x = \dfrac{15 + 3p}{10}$

The asymptotes intersect at the center.

Given: Center $= \left(h, k\right) = \left(2, 1\right)$

$\implies$ $\dfrac{15 + 3p}{10} = 2$ $\implies$ $p = \dfrac{5}{3}$

Substituting the value of $p$ in equation $(3)$, the equation of the second asymptote is

$x + 3y - 5 = 0$ $\;\;\; \cdots \; (4)$

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

$\left(3x - y - 5\right) \left(x + 3y - 5\right) = 0$ $\;\;\; \cdots \; (5)$

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

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

$\left(3x - y - 5\right) \left(x + 3y - 5\right) + \ell = 0$ $\;\;\; \cdots \; (6)$

Equation $(6)$ passes through the point $\left(1, -1\right)$.

$\therefore$ $\;$ We have

$\left(3 + 1 - 5\right) \left(1 - 3 - 5\right) + \ell = 0$ $\implies$ $\ell = -7$

$\therefore$ $\;$ The required equation of rectangular hyperbola is

$\left(3x - y - 5\right) \left(x + 3y - 5\right) - 7 = 0$

Analytical Geometry - Conics - Rectangular Hyperbola

A standard rectangular hyperbola has its vertices at $\left(5, 7\right)$ and $\left(-3, -1\right)$. Find its equation and asymptotes.

Let the required equation of the general rectangular hyperbola be

$\left(x - h\right) \left(y - k\right) = c^2$ $\;\;\; \cdots \; (1)$

The vertices of the rectangular hyperbola are $\left(5, 7\right)$ and $\left(-3, -1\right)$.

Let the center of the rectangular hyperbola be $\left(h, k\right)$.

The midpoint of the vertices of a rectangular hyperbola is its center.

$\therefore$ $\;$ We have, $\;$ $h = \dfrac{5 - 3}{2} = 1$ $\;$ and $\;$ $k = \dfrac{7 - 1}{2} = 3$

$\therefore$ $\;$ The center of the rectangular hyperbola is $\left(h, k\right) = \left(1, 3\right)$

Substituting the values of $h$ and $k$ in equation $(1)$ we have,

$\left(x -1\right) \left(y - 3\right) = c^2$ $\;\;\; \cdots \; (2)$

The vertices $\left(5, 7\right)$ and $\left(-3, -1\right)$ lie on the hyperbola.

$\therefore$ $\;$ We have from equation $(2)$,

$\left(5 - 1\right) \left(7 - 3\right) = c^2$ $\implies$ $c^2 = 16$

Substituting the value of $c^2$ in equation $(2)$, the required equation of rectangular hyperbola is

$\left(x -1\right) \left(y - 3\right) = 16$

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

$\therefore$ $\;$ The combined equation of the asymptotes is $\;$ $\left(x -1\right) \left(y - 3\right) = 0$

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

$x -1 = 0$ $\;$ and $\;$ $y - 3 = 0$

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equation of the rectangular hyperbola which which has for one of its asymptotes the line $x + 2y - 5 = 0$ and passes through the points $\left(6,0\right)$ and $\left(-3,0\right)$.

Given: One of the asymptotes of the rectangular hyperbola is

$x + 2y - 5 = 0$ $\;\;\; \cdots \; (1)$

$\therefore$ $\;$ Slope of the asymptote $= m_1 = \dfrac{-1}{2}$

The asymptotes of a rectangular hyperbola are at right angles to one another.

$\therefore$ $\;$ Slope of the second asymptote $= m_2 = \dfrac{-1}{m_1} = 2$

$\therefore$ $\;$ Let the equation of the second asymptote be

$y = 2x + k$ $\;$ where $k$ is a constant.

i.e. $\;$ $2x - y + k = 0$ $\;\;\; \cdots \; (2)$

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

$\left(x + 2y - 5\right) \left(2x - y + k\right) = 0$ $\;\;\; \cdots \; (3)$

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

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

$\left(x + 2y - 5\right) \left(2x - y + k\right) + \ell = 0$ $\;\;\; \cdots \; (4)$

Equation $(4)$ passes through the points $\left(6, 0\right)$ and $\left(-3, 0\right)$.

$\therefore$ $\;$ We have

$\left(6 + 0 - 5\right) \left(12 - 0 + k\right) + \ell = 0$ $\implies$ $k + \ell = - 12$ $\;\;\; \cdots \; (5a)$

and $\left(-3 + 0 - 5\right) \left(-6 - 0 + k\right) + \ell = 0$ $\implies$ $-8k + \ell = -48$ $\;\;\; \cdots \; (5b)$

Solving equations $(5a)$ and $(5b)$ simultaneously we have,

$k = 4$ $\;$ and $\;$ $\ell = -16$

Substituting the values of $k$ and $\ell$ in equation $(4)$, the equation of the required rectangular hyperbola is

$\left(x + 2y - 5\right) \left(2x - y + 4\right) - 16 = 0$

i.e. $\left(x + 2y - 5\right) \left(2x - y + 4\right) = 16$

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equation of the tangent and normal at $\left(-2, \dfrac{1}{4}\right)$ to the rectangular hyperbola $2xy - 2x - 8y - 1 = 0$

Equation of given rectangular hyperbola is

$2xy - 2x - 8y -1 = 0$

i.e. $\;$ $2xy - 2x - 8y = 1$

i.e. $\;$ $2xy - 2x - 8y + 8 = 9$

i.e. $\;$ $xy - x - 4y + 4 = \dfrac{9}{2}$

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

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

The standard equation of rectangular hyperbola with center at $\left(h, k\right)$ is:

$\left(x - h\right) \left(y - k\right) = c^2$ $\;\;\; \cdots \; (2)$

Comparing equations $(1)$ and $(2)$ we have

Center $= \left(h, k\right) = \left(4, 1\right)$ $\;$ and $\;$ $c^2 = \dfrac{9}{2}$

Equation of tangent to the rectangular hyperbola at the point $\left(x_1, y_1\right)$ is

$x \left(y_1 - k\right) + y \left(x_1 - h\right) = c^2 - hk + x_1y_1$

Given: Point $\left(x_1, y_1\right) = \left(-2, \dfrac{1}{4}\right)$

$\therefore$ $\;$ The required equation of tangent is

$x \left(\dfrac{1}{4} - 1\right) + y \left(-2 - 4\right) = \dfrac{9}{2} - \left(4 \times 1\right) + \left(-2 \times \dfrac{1}{4}\right)$

i.e. $\;$ $\dfrac{-3}{4}x - 6y = 0$

i.e. $\;$ $\dfrac{-x}{4}- 2y = 0$

i.e. $\;$ $x + 8y = 0$

Equation of normal to the rectangular hyperbola at the point $\left(x_1, y_1\right)$ is

$x \left(x_1 - h\right) - y \left(y_1 - k\right) = x_1 \left(x_1 - h\right) - y_1 \left(y_1 - k\right)$

$\therefore$ $\;$ The required equation of normal is

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

i.e. $\;$ $- 6x + \dfrac{3}{4}y = \dfrac{195}{16}$

i.e. $\;$ $- 2x + \dfrac{y}{4} = \dfrac{65}{16}$

i.e. $\;$ $32x - 4y + 65 = 0$

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equation of the standard rectangular hyperbola whose center is $\left(\dfrac{-1}{2}, \dfrac{-1}{2}\right)$ and which passes through the point $\left(1, \dfrac{1}{4}\right)$.

The equation of standard rectangular hyperbola with center at $\left(h, k\right)$ is:

$\left(x - h\right) \left(y - k\right) = c^2$ $\;\;\; \cdots \; (1)$

Given: Center $= \left(h, k\right) = \left(\dfrac{-1}{2}, \dfrac{-1}{2}\right)$

Putting the values of $h$ and $k$ in equation $(1)$, the required equation of rectangular hyperbola becomes

$\left(x + \dfrac{1}{2}\right) \left(y + \dfrac{1}{2}\right) = c^2$ $\;\;\; \cdots \; (2)$

Equation $(2)$ passes through the point $\left(1, \dfrac{1}{4}\right)$.

$\therefore$ $\;$ We have

$\left(1 + \dfrac{1}{2}\right) \left(\dfrac{1}{4} + \dfrac{1}{2}\right) = c^2$ $\implies$ $c^2 = \dfrac{9}{8}$

$\therefore$ $\;$ The required equation of the rectangular hyperbola [equation $(2)$] is

$\left(x + \dfrac{1}{2}\right) \left(y + \dfrac{1}{2}\right) = \dfrac{9}{8}$