Coordinate Geometry - Parabola

If the normal at $P \left(18, 12\right)$ to the parabola $y^2 = 8x$ cuts it again at $Q$, show that $\;$ $9PQ = 80 \sqrt{10}$

Equation of parabola: $\;$ $y^2 = 8x$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 8$ $\implies$ $a = 2$

Equation of normal to the parabola $\;$ $y^2 = 4ax$ $\;$ at $\;$ $P \left(x_1, y_1\right)$ $\;$ is

$y - y_1 = \left(\dfrac{-y_1}{2a}\right) \left(x - x_1\right)$

Given point: $\;$ $P \left(18, 12\right)$

$\therefore \;$ Equation of normal at $P \left(18, 12\right)$ to parabola $(1)$ is

$y - 12 = \left(\dfrac{-12}{2 \times 2}\right) \left(x - 18\right)$

i.e. $\;$ $y - 12 = -3x + 54$

i.e. $\;$ $3x + y - 66 = 0$ $\;\;\; \cdots \; (2)$

To determine the point $Q$ where the normal again meets the parabola, substitute $\;$ $y = 66 - 3x$ $\;$ from equation $(2)$ in equation $(1)$

$\therefore \;$ We have,

$\left(66 - 3x\right)^2 = 8x$

i.e. $\;$ $4356 + 9x^2 - 396x = 8x$

i.e. $\;$ $9x^2 - 404x + 4356 = 0$

Solving, we get, $\;$ $x = \dfrac{242}{9}$ $\;$ or $\;$ $x = 18$

When $\;$ $x = \dfrac{242}{9}$, $\;$ we have from equation $(2)$

$y = 66 - 3 \times \dfrac{242}{9} = \dfrac{-44}{3}$

$\therefore \;$ $Q = \left(\dfrac{242}{9}, \dfrac{-44}{3}\right)$

$\therefore \;$ $PQ = \sqrt{\left(18 - \dfrac{242}{9}\right)^2 + \left(12 + \dfrac{44}{3}\right)^2}$

i.e. $\;$ $PQ = \sqrt{\dfrac{6400}{81} + \dfrac{6400}{9}} = 80 \sqrt{\dfrac{1 + 9}{81}} = \dfrac{80 \sqrt{10}}{9}$

$\implies$ $9 PQ = 80 \sqrt{10}$

Hence proved.

Coordinate Geometry - Parabola

Find the equation of the normal to the parabola $y^2 = 4x$ which is perpendicular to the line $2x + 6y + 5 = 0$.

Equation of parabola: $\;$ $y^2 = 4x$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 4$ $\implies$ $a = 1$

Equation of line: $\;$ $2x + 6y + 5 = 0$ $\;\;\; \cdots \; (2)$

Slope of equation $(2)$ is $\;$ $m_1 = \dfrac{-2}{6} = \dfrac{-1}{3}$

The required normal is perpendicular to line $(2)$.

$\therefore \;$ Slope of normal $= m = \dfrac{-1}{m_1} = 3$

Let the equation of the required normal be $\;$ $y = mx - 2am - am^2$ $\;\;\; \cdots \; (3)$

Substituting the values of $a$ and $m$ in equation $(3)$ gives

$y = 3x - 2 \times 1 \times 3 - 1 \times 3^3$

i.e. $\;$ $3x - y - 33 = 0$ $\;\;\; \cdots \; (4)$

Equation $(4)$ is the equation of the required normal.

Coordinate Geometry - Parabola

Find the equation of the tangent to the parabola $y^2 = 16x$ which is parallel to the line $3x - 4y + 5 = 0$. Also find the point of contact.

Equation of parabola: $\;$ $y^2 = 16x$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 16$ $\implies$ $a = 4$

Equation of line: $\;$ $3x - 4y + 5 = 0$ $\;\;\; \cdots \; (2)$

Slope of equation $(2)$ is $\;$ $m = \dfrac{3}{4}$

The required tangent is parallel to line $(2)$.

$\therefore \;$ Slope of tangent $= m = \dfrac{3}{4}$

Let the equation of the required tangent be $\;$ $y = mx + \dfrac{a}{m}$ $\;\;\; \cdots \; (3)$

Substituting the values of $a$ and $m$ in equation $(3)$ gives

$y = \dfrac{3}{4} x + \dfrac{4}{3 / 4}$

i.e. $\;$ $y = \dfrac{3}{4} x + \dfrac{16}{3}$ $\;\;\; \cdots \; (4a)$

i.e. $\;$ $9x - 12y + 64 = 0$ $\;\;\; \cdots \; (4b)$

Equation $(4b)$ is the equation of the required tangent.

Substituting the value of $y$ from equation $(4a)$ in equation $(1)$ gives the point of contact.

$\therefore \;$ We have

$\left(\dfrac{3}{4} x + \dfrac{16}{3}\right)^2 = 16x$

i.e. $\;$ $\dfrac{9}{16} x^2 + \dfrac{256}{9} + 8x = 16x$

i.e. $\;$ $81 x^2 - 1152 x + 4096 = 0$

i.e. $\;$ $x = \dfrac{1152 \pm \sqrt{\left(-1152\right)^2 - 4 \times 81 \times 4096}}{2 \times 81}$

i.e. $\;$ $x = \dfrac{1152 \pm \sqrt{1327104 - 1327104}}{162}$

i.e. $\;$ $x = \dfrac{1152}{162} = \dfrac{64}{9}$

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

$y = \left(\dfrac{3}{4} \times \dfrac{64}{9}\right) + \dfrac{16}{3} = \dfrac{32}{3}$

$\therefore \;$ The point of contact is $\left(\dfrac{64}{9}, \dfrac{32}{3}\right)$.

Coordinate Geometry - Parabola

Find the value of $k$ so that the line $3x - y + k = 0$ may touch the parabola $y^2 = 24x$.

Given parabola: $\;$ $y^2 = 24x$ $\;\;\; \cdots \; (1)$

Given line: $\;$ $3x - y + k = 0$ $\;$ i.e. $\;$ $y = 3x + k$ $\;\;\; \cdots \; (2)$

Substituting the value of $y$ from equation $(2)$ in equation $(1)$ gives

$\left(3x + k\right)^2 = 24x$

i.e. $\;$ $9x^2 + 6kx + k^2 = 24x$

i.e. $\;$ $9x^2 + \left(6k - 24\right) x + k^2 = 0$ $\;\;\; \cdots \; (3)$

Since equation $(2)$ touches the parabola given by equation $(1)$, line $(2)$ meets the parabola in two coincident points.

i.e. $\;$ the discriminant of equation $(3)$ is equal to $0$ (zero).

i.e. $\;$ Discriminant $= \Delta = \left(6k - 24\right)^2 - 4 \times 9 \times k^2 = 0$

i.e. $\;$ $36 k^2 - 288 k + 576 - 36 k^2 = 0$

i.e. $\;$ $288 k = 576$

$\implies$ $k = 2$

Coordinate Geometry - Parabola

Show that the line $12y - 20x - 9 = 0$ touches the parabola $y^2 = 5x$.

Equation of parabola: $\;$ $y^2 = 5x$ $\;\;\; \cdots \; (1)$

Equation of line: $\;$ $12y - 20x - 9 = 0$ $\;$ i.e. $\;$ $y = \dfrac{20x + 9}{12}$ $\;\;\; \cdots \; (2)$

In view of equation $(2)$, equation $(1)$ becomes

$\left(\dfrac{5}{3} x + \dfrac{3}{4}\right)^2 = 5x$

i.e. $\;$ $\dfrac{25}{9} x^2 + \dfrac{9}{16} + \dfrac{5}{2} x = 5x$

i.e. $\;$ $\dfrac{25}{9} x^2 - \dfrac{5}{2} x + \dfrac{9}{16} = 0$ $\;\;\; \cdots \; (3)$

Discriminant of equation $(3)$ is

$\Delta = \left(\dfrac{-5}{2}\right)^2 - 4 \times \dfrac{25}{9} \times \dfrac{9}{16} = \dfrac{25}{4} - \dfrac{25}{4} = 0$

$\implies$ Roots of equation $(3)$ are equal.

$\implies$ Equation $(2)$ touches equation $(1)$

i.e. $\;$ equation $(2)$ is a tangent to the given parabola.

Coordinate Geometry - Parabola

Find the equations of the tangents to the parabola $y^2 + 12x = 0$, from the point $\left(3, 8\right)$.

Equation of parabola: $\;$ $y^2 + 12x = 0$ $\;$ i.e. $\;$ $y^2 = -12x$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = -12$ $\implies$ $a = -3$

Let the equation of the required tangent be

$y = mx + \dfrac{a}{m}$ $\;\;\; \cdots \; (2a)$

Substituting the value of $\;$ $a$ $\;$ in equation $(2a)$ gives

$y = mx - \dfrac{3}{m}$ $\;\;\; \cdots \; (2b)$

Given: $\;$ Equation $(2b)$ is drawn from the point $\left(3, 8\right)$,

$\implies$ $8 = 3m - \dfrac{3}{m}$

i.e. $\;$ $3m^2 - 8m - 3 = 0$

i.e. $\;$ $\left(3m + 1\right) \left(m - 3\right) = 0$

i.e. $\;$ $m = \dfrac{-1}{3}$ $\;$ or $\;$ $m = 3$

Substituing the values of $m$ in equation $(2b)$ gives

when $\;$ $m = \dfrac{-1}{3}$

$y = \dfrac{-1}{3} x - \dfrac{3}{-1/3}$

i.e. $\;$ $3y = -x + 27$

i.e. $\;$ $x + 3y = 27$ $\;\;\; \cdots \; (3a)$

when $\;$ $m = 3$

$y = 3x - \dfrac{3}{3}$

i.e. $\;$ $3x - y = 1$ $\;\;\; \cdots \; (3b)$

Equations $(3a)$ and $(3b)$ are the required equations of tangents.

Coordinate Geometry - Parabola

A tangent to the parabola $\;$ $y^2 = 16x$ $\;$ makes an angle of $\;$ $60^\circ$ $\;$ with the axis. Find the point of contact.

Equation of parabola: $\;$ $y^2 = 16x$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 16$ $\implies$ $a = 4$

Let slope of tangent $= m$

Since the tangent makes an angle of $60^\circ$ with the axis,

$\implies$ $m = \tan 60^\circ = \sqrt{3}$

Point of contact of the tangent with the parabola $= \left(\dfrac{a}{m^2}, \dfrac{2a}{m}\right) = \left(\dfrac{4}{\left(\sqrt{3}\right)^2}, \dfrac{2 \times 4}{\sqrt{3}}\right)$

$\therefore \;$ The point of contact $= \left(\dfrac{4}{3}, \dfrac{8}{\sqrt{3}}\right)$

Coordinate Geometry - Parabola

Find the tangent to the parabola $\;$ $y^2 = 16x$ $\;$ making an angle of $\;$ $45^\circ$ $\;$ with the $X$ axis.

Let the equation of the tangent be

$y = mx + \dfrac{a}{m}$ $\;\;\; \cdots \; (1)$ $\;\;\;$ where $m$ is the slope of the tangent

The tangent makes an angle of $45^\circ$ with the $X$ axis.

$\therefore \;$ $m = \tan 45^\circ = 1$

Given: $\;$ Equation of parabola: $\;\;\;$ $y^2 = 16x$ $\;\;\; \cdots \; (2)$

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

$4a = 16$ $\implies$ $a = 4$

Substituting the values of $m$ and $a$ in equation $(1)$ gives

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

i.e. $\;$ $y = x + 4$ $\;\;\; \cdots \; (3)$

Equation $(3)$ is the required equation of tangent.

Coordinate Geometry - Parabola

Find the equation of the parabola whose axis is parallel to the $X$ axis and the parabola passes through $\left(3, 3\right)$, $\left(6, 5\right)$ and $\left(6, -3\right)$.

The axis of the required parabola is parallel to the $X$ axis.

$\therefore \;$ Let the equation of the parabola be

$x = Ay^2 + By + C$ $\;\;\; \cdots \; (1)$

Since equation $(1)$ passes through the points $\left(3, 3\right)$, $\left(6, 5\right)$ and $\left(6, -3\right)$, we have

$3 = 9A + 3B + C$ $\;\;\; \cdots \; (2a)$

$6 = 25 A + 5B + C$ $\;\;\; \cdots \; (2b)$

$6 = 9A - 3B + C$ $\;\;\; \cdots \; (2c)$

Solving equations $(2a)$ and $(2c)$ simultaneously gives

$- 3 = 6 B$ $\implies$ $B = \dfrac{-1}{2}$

Solving equations $(2a)$ and $(2b)$ simultaneously gives

$3 = 16A + 2B$ $\;\;\; \cdots \; (3a)$

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

$3 = 16A + 2 \times \left(\dfrac{-1}{2}\right)$

i.e. $\;$ $4 = 16 A$ $\implies$ $A = \dfrac{1}{4}$

Substituting the values of $A$ and $B$ in equation $(2a)$ gives

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

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

Substituting the values of $A$, $B$ and $C$ in equation $(1)$ gives

$x = \dfrac{1}{4} y^2 - \dfrac{1}{2} y + \dfrac{9}{4}$

i.e. $\;$ $y^2 - 2y - 4x + 9 = 0$ $\;\;\; \cdots \; (4)$

Equation $(4)$ is the required equation of the parabola.

Coordinate Geometry - Parabola

Find the equation of the parabola whose vertex is at $\left(3, -2\right)$ and the focus at $\left(6, 2\right)$.

Given: $\;$ Vertex $= V = \left(3, -2\right)$, $\;\;\;$ Focus $= F = \left(6, 2\right)$

The axis is the line joining the vertex and the focus.

$\therefore \;$ Slope of axis $= \dfrac{2 + 2}{6 - 3} = \dfrac{4}{3}$

The directrix is the line perpendicular to the axis and passing through the vertex.

$\therefore \;$ Slope of directrix $= \dfrac{-3}{4}$

Let $\;$ $Z \left(x_1, y_1\right)$ $\;$ be the point of intersection of the axis and the directrix.

Then, vertex $\;$ $V \left(3, -2\right)$ $\;$ is the midpoint of the segment $\;$ $ZF$ $\;$ where $\;$ $Z = \left(x_1, y_1\right)$ $\;$ and $\;$ $F = \left(6, 2\right)$

$\therefore \;$ By midpoint formula,

$3 = \dfrac{x_1 + 6}{2}$ $\implies$ $x_1 = 0$

and $\;$ $-2 = \dfrac{y_1 + 2}{2}$ $\implies$ $y_1 = -6$

$\therefore \;$ $Z \left(x_1, y_1\right) = \left(0, -6\right)$

$\therefore \;$ Equation of directrix $\;$ (with slope $\dfrac{-3}{4}$ and passing through $Z$) $\;$ is

$y + 6 = \dfrac{-3}{4} \left(x - 0\right)$

i.e. $\;$ $3x + 4y + 24 = 0$

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

Then, by definition,

distance of $P$ from focus $=$ distance of $P$ from the directrix

i.e. $\;$ $\sqrt{\left(x - 6\right)^2 + \left(y - 2\right)^2} = \left|\dfrac{3x + 4y + 24}{\sqrt{3^2 + 4^2}}\right|$

i.e. $\;$ $x^2 + y^2 - 12x - 4y + 36 + 4 = \dfrac{9x^2 + 16y^2 + 576 + 24xy + 144x + 192y}{5}$

i.e. $\;$ $5x^2 + 5y^2 - 60x - 20y + 200 = 9x^2 + 16y^2 + 24xy + 144x + 192y + 576$

i.e. $\;$ $4x^2 + 11y^2 - 24xy + 204x + 212y + 376 = 0$ $\;\;\; \cdots \; (1)$

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

Coordinate Geometry - Parabola

Find the equation of the parabola whose focus is at $\left(-2, -1\right)$ and the latus rectum joins the points $\left(-2, 2\right)$ and $\left(-2, -4\right)$.

Given: $\;$ Latus rectum joins the points $\;$ $\left(-2, 2\right)$ $\;$ and $\;$ $\left(-2, -4\right)$

Length of latus rectus $= \sqrt{\left(-2 + 2\right)^2 + \left(2 + 4\right)^2} = 6$

But length of latus rectum $= 4a$

i.e. $\;$ $4a = 6$ $\implies$ $a = \dfrac{6}{4} = \dfrac{3}{2}$

Given: $\;$ Focus $= \left(-2, -1\right)$

Let the vertex of the required parabola be $= \left(\alpha, -1\right)$

Then,

Distance between focus and vertex $= a$

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

i.e. $\;$ $\alpha + 2 = \dfrac{3}{2}$ $\implies$ $\alpha = \dfrac{-1}{2}$

$\therefore \;$ The vertex of the required parabola is $= \left(h, k\right) = \left(\dfrac{-1}{2}, -1\right)$

Let the equation of the required parabola be

$\left(y - k\right)^2 = \pm 4 a \left(x - h\right)$

i.e. $\;$ $\left(y + 1\right)^2 = \pm 6 \left(x + \dfrac{1}{2}\right)$

i.e. $\;$ $y^2 + 2y + 1 = 6x + 3$ $\;\;$ or $\;\;$ $y^2 + 2y + 1 = -6x - 3$

i.e. $\;$ $y^2 + 2y - 6x -2 = 0$ $\;\;\; \cdots \; (1a)$

or $\;$ $y^2 + 2y + 6x + 4 = 0$ $\;\;\; \cdots \; (1b)$

Equations $(1a)$ and $(1b)$ are the required equations of the parabola.

Coordinate Geometry - Parabola

Find the equation of the parabola whose focus is at $\left(1,1\right)$ and the directrix is $x - y = 3$.

Given: $\;$ Equation of directrix of parabola is $\;\;$ $x - y = 3$ $\;\;$ i.e. $\;$ $x - y -3 = 0$

Given: $\;$ Focus of parabola is at $\;$ $\left(1, 1\right)$

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

Then, by definition of parabola,

Distance of $P$ from the focus $=$ distance of point $P$ from the directrix

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

i.e. $\;$ $x^2 - 2x + 1 + y^2 - 2y + 1 = \dfrac{x^2 + y^2 + 9 - 2xy - 6x + 6y}{2}$

i.e. $\;$ $2x^2 - 4x + 2y^2 - 4y + 4 = x^2 + y^2 - 6x + 6y - 2xy + 9$

i.e. $\;$ $x^2 + y^2 + 2xy + 2x - 10 y - 5 = 0$ $\;\;\; \cdots \; (1)$

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

Coordinate Geometry - Parabola

Find the vertex, axis, focus and directrix of the parabola $\;$ $5y^2 - 20y - 16x + 96 = 0$.

Equation of given parabola: $\;$ $5y^2 - 20y - 16x + 96 = 0$

i.e. $\;$ $y^2 - 4y - \dfrac{16}{5}x + \dfrac{96}{5} = 0$

i.e. $\;$ $\left(y^2 - 4y + 4\right) = \dfrac{16}{5} x - \dfrac{96}{5} + 4$

i.e. $\;$ $\left(y - 2\right)^2 = \dfrac{16}{5} x - \dfrac{76}{5}$

i.e. $\;$ $\left(y - 2\right)^2 = \dfrac{16}{5} \left(x - \dfrac{76}{16}\right)$

i.e. $\;$ $\left(y - 2\right)^2 = \dfrac{16}{5} \left(x - \dfrac{19}{4}\right)$ $\;\;\; \cdots \; (1)$

To shift the origin to the point $\left(\dfrac{19}{4}, 2\right)$,

let $\;$ $x - \dfrac{19}{4} = X$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $y - 2 = Y$ $\;\;\; \cdots \; (2b)$

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

$Y^2 = \dfrac{16}{5} X$ $\;\;\; \cdots \; (3)$

1. Vertex of equation $(3)$ is $\;$ $\left(0, 0\right)$
   
   i.e. $\;$ $X = 0, \;\;\; Y = 0$
   
   When $\;$ $X = 0$
   
   $\implies$ $x - \dfrac{19}{4} = 0$ $\;$ i.e. $\;$ $x = \dfrac{19}{4}$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = 0$
   
   $\implies$ $y - 2 = 0$ $\;$ i.e. $\;$ $y = 2$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Vertex of equation $(1)$ is $\;$ $\left(\dfrac{19}{4}, 2\right)$
   
   
2. Axis of equation $(3)$ is $\;\;$ $Y = 0$
   
   $\therefore \;$ Axis of equation $(1)$ is $\;\;\;$ $y - 2 = 0$ $\;\;\;$ [by equation $(2b)$]
   
   i.e. $\;$ $y = 2$
   
   
3. Comparing equation $(3)$ with the standard equation of parabola $\;\;$ $Y^2 = 4a X$ $\;\;$ gives
   
   $4a = \dfrac{16}{5}$ $\implies$ $a = \dfrac{4}{5}$
   
   $\therefore \;$ Focus of equation $(3)$ is $\;\;\;$ $\left(a, 0\right) = \left(\dfrac{4}{5}, 0\right)$
   
   $\implies$ $X = \dfrac{4}{5}, \;\;\; Y = 0$
   
   When $\;$ $X = \dfrac{4}{5}$
   
   $\implies$ $x - \dfrac{19}{4} = \dfrac{4}{5}$ $\;$ i.e. $\;$ $x = \dfrac{111}{20}$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = 0$
   
   $\implies$ $y - 2 = 0$ $\;$ i.e. $\;$ $y = 2$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Focus of equation $(1)$ is $\;\;\;$ $\left(\dfrac{111}{20}, 2\right)$
   
   
4. Directrix of equation $(3)$ is $\;\;\;$ $X = -a$
   
   i.e. $\;$ $X = \dfrac{-4}{5}$
   
   i.e. $x - \dfrac{19}{4} = \dfrac{-4}{5}$ $\;\;$ i.e. $\;\;$ $x = \dfrac{79}{20}$ $\;\;\;$ [by equation $(2a)$]
   
   $\therefore \;$ Directrix of equation $(1)$ is $\;\;\;$ $20 x - 79 = 0$

Coordinate Geometry - Parabola

Find the vertex, axis, focus and directrix of the parabola $\;$ $x^2 + 2gx + 2fy + c = 0$.

Equation of given parabola: $\;$ $x^2 + 2gx + 2fy +c = 0$

i.e. $\;$ $\left(x^2 + 2gx+ g^2\right) + 2fy + c - g^2 = 0$

i.e. $\;$ $\left(x + g\right)^2 = -2fy + g^2 - c$

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

To shift the origin to the point $\left(-g, \dfrac{g^2 - c}{2f}\right)$,

let $\;$ $x + g = X$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $y - \dfrac{g^2 - c}{2f} = Y$ $\;\;\; \cdots \; (2b)$

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

$X^2 = -2fY$ $\;\;\; \cdots \; (3)$

1. Vertex of equation $(3)$ is $\;$ $\left(0, 0\right)$
   
   i.e. $\;$ $X = 0, \;\;\; Y = 0$
   
   When $\;$ $X = 0$
   
   $\implies$ $x + g = 0$ $\;$ i.e. $\;$ $x = -g$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = 0$
   
   $\implies$ $y - \dfrac{g^2 - c}{2f} = 0$ $\;$ i.e. $\;$ $y = \dfrac{g^2 - c}{2f}$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Vertex of equation $(1)$ is $\;$ $\left(-g, \dfrac{g^2 - c}{2f}\right)$
   
   
2. Axis of equation $(3)$ is $\;\;$ $X = 0$
   
   $\therefore \;$ Axis of equation $(1)$ is $\;\;\;$ $x + g = 0$ $\;\;\;$ [by equation $(2a)$]
   
   i.e. $\;$ $x = -g$
   
   
3. Comparing equation $(3)$ with the standard equation of parabola $\;\;$ $X^2 = -4a Y$ $\;\;$ gives
   
   $4a = 2f$ $\implies$ $a = \dfrac{f}{2}$
   
   $\therefore \;$ Focus of equation $(3)$ is $\;\;\;$ $\left(0, -a\right) = \left(0, \dfrac{-f}{2}\right)$
   
   $\implies$ $X = 0, \;\;\; Y = \dfrac{-f}{2}$
   
   When $\;$ $X = 0$
   
   $\implies$ $x + g = 0$ $\;$ i.e. $\;$ $x = -g$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = \dfrac{-f}{2}$
   
   $\implies$ $y - \dfrac{g^2 - c}{2f} = \dfrac{-f}{2}$ $\;$ i.e. $\;$ $y = \dfrac{g^2 - f^2 - c}{2f}$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Focus of equation $(1)$ is $\;\;\;$ $\left(-g, \dfrac{g^2 - f^2 - c}{2f}\right)$
   
   
4. Directrix of equation $(3)$ is $\;\;\;$ $Y = a$
   
   i.e. $\;$ $Y = \dfrac{f}{2}$
   
   i.e. $y - \dfrac{g^2 - c}{2f} = \dfrac{f}{2}$ $\;\;\;$ [by equation $(2b)$]
   
   i.e. $\;$ $y= \dfrac{g^2 - f^2 - c}{2f}$
   
   $\therefore \;$ Directrix of equation $(1)$ is $\;\;\;$ $y = \dfrac{g^2 - f^2 - c}{2f}$

Coordinate Geometry - Parabola

Find the vertex, axis, focus and directrix of the parabola $\;$ $\left(x - 4\right)^2 = -8y + 5$.

Equation of given parabola: $\;$ $\left(x - 4\right)^2 = -8y + 5$

i.e. $\;$ $\left(x - 4\right)^2 = -8 \left(y - \dfrac{5}{8}\right)$ $\;\;\; \cdots \; (1)$

To shift the origin to the point $\left(4, \dfrac{5}{8}\right)$,

let $\;$ $x - 4 = X$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $y - \dfrac{5}{8} = Y$ $\;\;\; \cdots \; (2b)$

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

$X^2 = -8Y$ $\;\;\; \cdots \; (3)$

1. Vertex of equation $(3)$ is $\;$ $\left(0, 0\right)$
   
   i.e. $\;$ $X = 0, \;\;\; Y = 0$
   
   When $\;$ $X = 0$
   
   $\implies$ $x - 4 = 0$ $\;$ i.e. $\;$ $x = 4$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = 0$
   
   $\implies$ $y - \dfrac{5}{8} = 0$ $\;$ i.e. $\;$ $y = \dfrac{5}{8}$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Vertex of equation $(1)$ is $\;$ $\left(4, \dfrac{5}{8}\right)$
   
   
2. Axis of equation $(3)$ is $\;\;$ $X = 0$
   
   $\therefore \;$ Axis of equation $(1)$ is $\;\;\;$ $x - 4 = 0$ $\;\;\;$ [by equation $(2a)$]
   
   i.e. $\;$ $x = 4$
   
   
3. Comparing equation $(3)$ with the standard equation of parabola $\;\;$ $X^2 = -4a Y$ $\;\;$ gives
   
   $4a = 8$ $\implies$ $a = 2$
   
   $\therefore \;$ Focus of equation $(3)$ is $\;\;\;$ $\left(0, -a\right) = \left(0, -2\right)$
   
   $\implies$ $X = 0, \;\;\; Y = -2$
   
   When $\;$ $X = 0$
   
   $\implies$ $x - 4 = 0$ $\;$ i.e. $\;$ $x = 4$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = -2$
   
   $\implies$ $y - \dfrac{5}{8} = -2$ $\;$ i.e. $\;$ $y = \dfrac{-11}{8}$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Focus of equation $(1)$ is $\;\;\;$ $\left(4, \dfrac{-11}{8}\right)$
   
   
4. Directrix of equation $(3)$ is $\;\;\;$ $Y = a$
   
   i.e. $\;$ $Y = 2$
   
   i.e. $y - \dfrac{5}{8} = 2$ $\;\;\;$ [by equation $(2b)$]
   
   i.e. $\;$ $y= \dfrac{21}{8}$
   
   $\therefore \;$ Directrix of equation $(1)$ is $\;\;\;$ $8y = 21$

Coordinate Geometry - Parabola

Find the vertex, axis, focus and directrix of the parabola $\;$ $\left(y + 3\right)^2 = 4x + 4$.

Equation of given parabola: $\;$ $\left(y + 3\right)^2 = 4x + 4$

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

To shift the origin to the point $\left(-1, -3\right)$,

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

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

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

1. Vertex of equation $(3)$ is $\;$ $\left(0, 0\right)$
   
   i.e. $\;$ $X = 0, \;\;\; Y = 0$
   
   When $\;$ $X = 0$
   
   $\implies$ $x + 1 = 0$ $\;$ i.e. $\;$ $x = -1$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = 0$
   
   $\implies$ $y + 3 = 0$ $\;$ i.e. $\;$ $y = -3$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Vertex of equation $(1)$ is $\;$ $\left(-1, -3\right)$
   
   
2. Axis of equation $(3)$ is $\;\;$ $Y = 0$
   
   $\therefore \;$ Axis of equation $(1)$ is $\;\;\;$ $y + 3 = 0$ $\;\;\;$ [by equation $(2b)$]
   
   i.e. $\;$ $y = -3$
   
   
3. Comparing equation $(3)$ with the standard equation of parabola $\;\;$ $Y^2 = 4a X$ $\;\;$ gives
   
   $a = 1$
   
   $\therefore \;$ Focus of equation $(3)$ is $\;\;\;$ $\left(a, 0\right) = \left(1, 0\right)$
   
   $\implies$ $X = 1, \;\;\; Y = 0$
   
   When $\;$ $X = 1$
   
   $\implies$ $x + 1 = 1$ $\;$ i.e. $\;$ $x = 0$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = 0$
   
   $\implies$ $y + 3 = 0$ $\;$ i.e. $\;$ $y = -3$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Focus of equation $(1)$ is $\;\;\;$ $\left(0, -3\right)$
   
   
4. Directrix of equation $(3)$ is $\;\;\;$ $X = -a$
   
   i.e. $\;$ $X = -1$
   
   i.e. $x + 1 = -1$ $\;\;\;$ [by equation $(2a)$]
   
   $\therefore \;$ Directrix of equation $(1)$ is $\;\;\;$ $x + 2 = 0$

Analytical Geometry - Conics - Parabola

A cable of a suspension bridge is in the form of a parabola whose span is 40 m. The roadway is 5 m below the lowest point of the cable. If an extra support is provided across the cable 30 m above the ground level, find the length of the support if the height of the pillars are 55 m.

Take the lowest point on the suspension bridge as the vertex and let it be at the origin.

$\therefore$ $\;$ Vertex $= V = \left(0,0\right)$

Let $\;$ $RR'$ $\;$ be the roadway.

$PQ$ $\;$ is the span of the bridge $= 40$ m

Let extra support be provided at the points $AA'$.

$BA =$ height of point $A$ from the roadway $= 30$ m

$B'A' =$ height of point $A'$ from the roadway $= 30$ m

$PP'$, $\;$ $QQ' =$ pillars of height $55$ m

$\because$ $\;$ $PQ = 40$ m $\implies$ $VD' = 20$ m

$\therefore$ $\;$ Coordinates of $Q = \left(20, 50\right)$

Equation of the parabola is $\;\;$ $x^2 = 4ay$ $\;\;\; \cdots \; (1)$

$\because$ $\;$ Point $Q$ lies on equation $(1)$, we have

$\left(20\right)^2 = 4 \times a \times 50$ $\implies$ $a = 2$

Substituting the value of $a$ in equation $(1)$ gives the equation of the parabola as

$x^2 = 8y$ $\;\;\; \cdots \; (2)$

Let $VC' = \ell$ m.

From the figure, $\;$ $A'C' = 25$ m

Then the coordinates of point $A'$ are $\left(\ell, 25\right)$

$\because$ $\;$ Point $A'$ lies on the parabola given by equation $(2)$, we have

$\ell^2 = 8 \times 25 = 200$ $\implies$ $\ell = 10 \sqrt{2}$

$\therefore$ $\;$ From the figure, $AA' = 2 \ell = 20 \sqrt{2}$

$\therefore$ $\;$ Length of support $= 20 \sqrt{2}$ m

Analytical Geometry - Conics - Parabola

The focus of a parabolic mirror is at a distance of 8 cm from its center (vertex). If the mirror is 25 cm deep, find the diameter of the mirror.

Let the vertex of the parabolic mirror be at the origin i.e. $V \left(0,0\right)$

Let the equation of the parabolic mirror be: $\;\;$ $y^2 = 4ax$ $\;\;\; \cdots \; (1)$

Given: $\;$ Focus is at a distance of $8$ cm from the center.

$\therefore$ $\;$ $F = \left(a, 0\right) = \left(8, 0\right)$ $\;\;$ $\implies$ $a = 8$ $\;\;\; \cdots \; (2)$

Let the diameter of the parabolic mirror be $AB$.

Depth of the mirror $= 25$ cm

Let $p$ be the radius of the parabolic mirror.

Then, $\;$ $A = \left(25, p\right)$ $\;\;\; \cdots \; (3)$

Now $A$ lies on the parabola.

$\therefore$ $\;$ We have from equations $(1)$, $(2)$ and $(3)$,

$p^2 = 4 \times 8 \times 25 = 800$ $\;\;$ $\implies$ $p = 20 \sqrt{2}$

$\therefore$ $\;$ Diameter $AB = 2p = 40 \sqrt{2}$ cm

Analytical Geometry - Conics - Parabola

Find the axis, vertex, focus, equation of directrix, latus rectum and length of the latus rectum for the parabola $y^2 + 8x - 6y + 1 = 0$. Also sketch its graph.

Given equation of parabola is: $\;\;$ $y^2 + 8x - 6y + 1 = 0$

i.e. $y^2 - 6y = - 8x - 1$

i.e. $y^2 - 6y + 9 = - 8x - 1 + 9$

i.e. $\left(y - 3\right)^2 = - 8 x + 8$

i.e. $\left(y - 3\right)^2 = - 8 \left(x - 1\right)$ $\;\;\; \cdots \; (1)$

Let $X = x - 1$ $\;$ and $\;$ $Y = y - 3$

Then equation $(1)$ can be written as

$Y^2 = - 8 X$

i.e. $\left(Y - 0\right)^2 = - 4 \times 2 \times \left(X - 0\right)$ $\;\;\; \cdots \; (2)$

Comparing equation $(2)$ with the standard equation $\;\;$ $\left(Y - k\right)^2 = - 4 \times a \times \left(X - h\right)$ $\;$ gives

$\left(h, k\right) = \left(0, 0\right)$ $\;$ and $\;$ $a = 2$

$\therefore$ $\;$ For the given parabola,

	Referred to X, Y	Referred to x, y $x = X + 1$; $\;$ $y = Y + 3$
Axis of symmetry	X axis $\;$ $\left(Y = 0\right)$	$Y = 0 \implies y = 3$ i.e. $\;$ $y - 3 = 0$
Vertex	$V = \left(h, k\right) = \left(0,0\right)$	$X = 0 \implies x = 1$ $Y = 0 \implies y = 3$ $\therefore$ $\;$ $V = \left(1,3\right)$
Focus	$F = \left(-a, 0\right) = \left(-2, 0\right)$	$X = -2 \implies x = -1$ $Y = 0 \implies y = 3$ $\therefore$ $\;$ $F = \left(-1, 3\right)$
Equation of directrix	$X = a$ i.e. $\;$ $X = 2$	$X = 2 \implies x = 3$ i.e. $\;$ $x - 3 = 0$
Equation of latus rectum	$X = -a$ i.e. $\;$ $X = -2$	$X = -2 \implies x = -1$ i.e. $\;$ $x + 1 = 0$
Length of latus rectum	$4a = 4 \times 2 = 8$	$8$

Analytical Geometry - Conics - Parabola

Find the axis, vertex, focus, equation of directrix, latus rectum and length of the latus rectum for the parabola $\left(x - 4\right)^2 = 4 \left(y + 2\right)$. Also sketch its graph.

Given equation of parabola is: $\;\;$ $\left(x - 4\right)^2 = 4 \left(y + 2\right)$ $\;\;\; \cdots \; (1a)$

Let $\;$ $X = x - 4$ $\;$ and $\;$ $Y = y + 2$

Then equation $(1a)$ can be written as: $\;\;$ $X^2 = 4 Y$

i.e. $\left(X - 0\right)^2 = 4 \times 1 \times \left(Y - 0\right)$ $\;\;\; \cdots \; (1b)$

Comparing equation $(1b)$ with the standard equation: $\;\;$ $\left(X - h\right)^2 = 4 a \left(Y - k\right)$ $\;$ gives

$\left(h, k\right) = \left(0, 0\right)$; $\;\;$ $a = 1$

$\therefore$ $\;$ For the given parabola,