Analytical Geometry - Conics - Tangents and Normals

Find the equations of tangent and normal to the parabola $y^2 = 8x$ at $t = \dfrac{1}{2}$.

Given: Equation of parabola is $\;$ $y^2 = 8x$

Comparing with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 8 \implies a = 2$

Equation of tangent to a parabola at a point $t$ is $\;$ $yt = x+ at^2$

Equation of normal to a parabola at a point $t$ is $\;$ $y + tx = 2at + at^3$

Given: $\;$ $t = \dfrac{1}{2}$

$\therefore$ $\;$ Required equation of tangent is:

$y \times \dfrac{1}{2} = x + 2 \times \dfrac{1}{4}$

i.e. $\;$ $2x - y + 1 = 0$

Required equation of normal is:

$y + \dfrac{x}{2} = 2 \times 2 \times \dfrac{1}{2} + 2 \times \dfrac{1}{8}$

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

i.e. $\;$ $2x + 4y - 9 = 0$

Analytical Geometry - Conics - Tangents and Normals

Find the equations of tangent and normal to the hyperbola $9x^2 - 5y^2 = 31$ at $\left(2,-1\right)$.

Equation of given hyperbola: $\;$ $9x^2 - 5y^2 = 31$

i.e. $\;$ $\dfrac{x^2}{31 / 9} - \dfrac{y^2}{31 / 5} = 1$ $\;\;\; \cdots \; (1)$

The transverse axis of the hyperbola 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

$a^2 = \dfrac{31}{9}$ $\;$ and $\;$ $b^2 = \dfrac{31}{5}$

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

$\dfrac{xx_1}{a^2} - \dfrac{yy_1}{b^2} = 1$

Given: $\left(x_1, y_1\right) = \left(2, -1\right)$

$\therefore$ $\;$ Required equation of tangent is

$\dfrac{2x}{31 / 9} + \dfrac{y}{31 / 5} = 1$

i.e. $\;$ $18x + 5y = 31$

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

$\dfrac{a^2 x}{x_1} + \dfrac{b^2 y}{y_1} = a^2 + b^2$

$\therefore$ $\;$ Required equation of normal is

$\dfrac{31 x}{9 \times 2} + \dfrac{31 y}{5 \times \left(-1\right)} = \dfrac{31}{9} + \dfrac{31}{5}$

i.e. $\;$ $\dfrac{x}{18} - \dfrac{y}{5} = \dfrac{14}{45}$

i.e. $\;$ $5x - 18 y - 28 = 0$

Analytical Geometry - Conics - Tangents and Normals

Find the equations of tangent and normal to the parabola $x^2 + 2x - 4y + 4 = 0$ at $\left(0,1\right)$.

Equation of given parabola is: $\;$ $x^2 + 2x - 4y + 4 = 0$

i.e. $\;$ $x^2 + 2x = 4y - 4$

i.e. $\;$ $x^2 + 2x +1 = 4y - 4 + 1$

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

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

Let $\;$ $X = x + 1$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $Y = y - \dfrac{3}{4}$ $\;\;\; \cdots \; (2b)$

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

$X^2 = 4Y$ $\;\;\; \cdots \; (3)$

Comparing equation $(3)$ with the standard equation of parabola $\;$ $X^2 = 4a Y$ $\;$ gives $\;$ $a = 1$

Given: Point $\left(x_1, y_1\right) = \left(0, 1\right)$

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

when $x_1 = 0 \implies X_1 = 0 + 1 = 1$

and when $y_1 = 1 \implies Y_1 = 1 - \dfrac{3}{4} = \dfrac{1}{4}$

$\therefore$ $\;$ $\left(X_1, Y_1\right) = \left(1, \dfrac{1}{4}\right)$

Equation of tangent at $\left(X_1, Y_1\right)$ to the parabola given by equation $(3)$ is

$XX_1 = 2a \left(Y + Y_1\right)$

$\therefore$ $\;$ Equation of tangent to the parabola given by equation $(3)$ at the point $\left(X_1, Y_1\right)$ is

$1 \times X = 2 \times 1 \times \left(Y + \dfrac{1}{4}\right)$

i.e. $\;$ $X = 2Y + \dfrac{1}{2}$ $\;\;\; \cdots \; (4)$

Equation of normal at $\left(X_1, Y_1\right)$ to the parabola given by equation $(3)$ is

$YX_1 + 2a X = X_1 Y_1 + 2a X_1$

i.e. $\;$ $1 \times Y + 2 \times 1 \times X = 1 \times \dfrac{1}{4} + 2 \times 1 \times 1$

i.e. $\;$ $Y + 2X = \dfrac{9}{4}$ $\;\;\; \cdots \; (5)$

Substituting for $X$ and $Y$ from equations $(2a)$ and $(2b)$ in equation $(4)$, the equation of tangent to parabola given by equation $(1)$ is

$x +1 = 2 \left(y - \dfrac{3}{4}\right) + \dfrac{1}{2}$

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

i.e. $\;$ $x - 2y + 2 = 0$

Substituting for $X$ and $Y$ from equations $(2a)$ and $(2b)$ in equation $(5)$, the equation of normal to the parabola given by equation $(1)$ is

$y - \dfrac{3}{4} + 2 \times \left(x + 1\right) = \dfrac{9}{4}$

i.e. $\;$ $2x + y - \dfrac{3}{4} + 2 - \dfrac{9}{4} = 0$

i.e. $\;$ $2x + y - 1 = 0$

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	$2b = 2 \times 4 = 8$	$8$