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 $\;\;\;$ $x + 2 = 0$

Coordinate Geometry - Circle

Find the equation of the circle whose center is $\left(5, \dfrac{\pi}{3}\right)$ and radius $7$.

Given: $\;\;$ Center of the circle $= \left(R, \phi\right) = \left(5, \dfrac{\pi}{3}\right)$

i.e. $\;$ $R = 5, \;\; \phi = \dfrac{\pi}{3}$

Radius of the circle $= a = 7$

Let the equation of the required circle be

$R^2 + r^2 - 2 R r \cos \left(\theta - \phi\right) = a^2$

Substituting the values of $R$, $\phi$ and $a$, we get

$25 + r^2 - 2 \times 5 \times r \cos \left(\theta - \dfrac{\pi}{3}\right) = 49$

i.e. $\;$ $r^2 - 10r \left(\cos \theta \cos \dfrac{\pi}{3} + \sin \theta \sin \dfrac{\pi}{3}\right) - 24 = 0$

i.e. $\;$ $r^2 - 10r \left(\dfrac{1}{2} \cos \theta + \dfrac{\sqrt{3}}{2} \sin \theta\right) - 24 = 0$

i.e. $\;$ $r^2 - 5r \left(\cos \theta + \sqrt{3} \sin \theta\right) - 24 = 0$

Coordinate Geometry - Circle

Find the center and radius of the circle $\;$ $r^2 - 2r \left(24 \cos \theta - 7 \sin \theta\right) + 141 = 0$

Equation of given circle: $\;$ $r^2 - 2r \left(24 \cos \theta - 7 \sin \theta\right) + 141 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $\left(R, \phi\right)$ $\;$ be the polar coordinates of the center and the radius be $= a$.

Then equation $(1)$ should be identical with

$r^2 - 2R r \cos \left(\theta - \phi\right) + R^2 - a^2 = 0$

i.e. $\;$ $r^2 - 2R r \left(\cos \theta \cos \phi + \sin \theta \sin \phi\right) + R^2 - a^2 = 0$ $\;\;\; \cdots \; (2)$

$\therefore \;$ $-4r \cos \theta = -2 R r \cos \theta \cos \phi$

i.e. $\;$ $R \cos \phi = 24$ $\;\;\; \cdots \; (3a)$;

$14 r \sin \theta = -2 R r \sin \theta \sin \phi$

i.e. $\;$ $R \sin \phi = -7$ $\;\;\; \cdots \; (3b)$

and $\;$ $R^2 - a^2 = 141$ $\;\;\; \cdots \; (3c)$

From equations $(3a)$ and $(3b)$

$\tan \phi = \dfrac{-7}{24}$

$\implies$ $\phi = \tan^{-1} \left(\dfrac{-7}{24}\right)$ $\;\;\; \cdots \; (4a)$

and $\;$ $R^2 \cos^2 \phi = 576, \;\; R^2 \sin^2 \phi = 49$

i.e. $\;$ $R^2 \left(\sin^2 \phi + \cos^2 \phi\right) = R^2 = 579 + 46 = 625$

$\implies$ $R = 25$

Substituting the value of $R^2$ in equation $(3c)$ gives

$625 - a^2 = 141$

i.e. $\;$ $a^2 = 484$ $\implies$ $a = 22$ $\;\;\; \cdots \; (4b)$

$\therefore \;$ The center of the circle is $\;$ $\left(25, \tan^{-1} \left(\dfrac{-7}{24}\right)\right)$ $\;$ and the radius is $\;$ $22$.

Coordinate Geometry - Circle

Find the center and radius of the circle $\;$ $r^2 - 4r \left(\sqrt{3} \cos \theta + \sin \theta\right) + 7 = 0$

Equation of given circle: $\;$ $r^2 - 4r \left(\sqrt{3} \cos \theta + \sin \theta\right) + 7 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $\left(R, \phi\right)$ $\;$ be the polar coordinates of the center and the radius be $= a$.

Then equation $(1)$ should be identical with

$r^2 - 2R r \cos \left(\theta - \phi\right) + R^2 - a^2 = 0$

i.e. $\;$ $r^2 - 2R r \left(\cos \theta \cos \phi + \sin \theta \sin \phi\right) + R^2 - a^2 = 0$

$\therefore \;$ $4r \times \sqrt{3} \cos \theta = 2 R r \cos \theta \cos \phi$

i.e. $\;$ $R \cos \phi = 2 \sqrt{3}$ $\;\;\; \cdots \; (3a)$;

$4 r \sin \theta = 2 R r \sin \theta \sin \phi$

i.e. $\;$ $R \sin \phi = 2$ $\;\;\; \cdots \; (3b)$

and $\;$ $R^2 - a^2 = 7$ $\;\;\; \cdots \; (3c)$

From equations $(3a)$ and $(3b)$

$\tan \phi = \dfrac{1}{\sqrt{3}}$

$\implies$ $\phi = \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6}$ $\;\;\; \cdots \; (4a)$

and $\;$ $R^2 \cos^2 \phi = 12, \;\; R^2 \sin^2 \phi = 4$

i.e. $\;$ $R^2 \left(\sin^2 \phi + \cos^2 \phi\right) = R^2 = 12 + 4 = 16$

$\implies$ $R = 4$

Substituting the value of $R^2$ in equation $(3c)$ gives

$16 - a^2 = 7$

i.e. $\;$ $a^2 = 9$ $\implies$ $a = 3$ $\;\;\; \cdots \; (4b)$

$\therefore \;$ The center of the circle is $\;$ $\left(4, \dfrac{\pi}{6}\right)$ $\;$ and the radius is $\;$ $3$.

Coordinate Geometry - Circle

Determine the limiting points of the system of circles $\;$ $x^2 + y^2 -3x - 9y + 45 - \lambda \left(x^2 + y^2 -2x -6y + 30\right) = 0$

Given equation of system of circles

$x^2 + y^2 -3x - 9y + 45 - \lambda \left(x^2 + y^2 -2x -6y + 30\right) = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ can be written as

$x^2 \left(1 - \lambda\right) + y^2 \left(1 - \lambda\right) + \left(2 \lambda - 3\right) x + \left(6 \lambda - 9\right) y + \left(45 - 30 \lambda\right) = 0$

i.e. $\;$ $x^2 + y^2 + \left(\dfrac{2 \lambda - 3}{1 - \lambda}\right) x + \left(\dfrac{6 \lambda - 9}{1 - \lambda}\right) y + \dfrac{45 - 30 \lambda}{1 - \lambda} = 0$ $\;\;\; \cdots \; (2)$

Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives

$g = \dfrac{2 \lambda - 3}{2 - 2 \lambda}, \;\; f = \dfrac{6 \lambda - 9}{2 - 2 \lambda}, \;\; c = \dfrac{45 - 30 \lambda}{1 - \lambda}$

Center of equation $(2)$ $= \left(-g, -f\right) = \left(\dfrac{3 - 2 \lambda}{2 - 2 \lambda}, \dfrac{9 - 6 \lambda}{2 - 2\lambda}\right)$ $\;\;\; \cdots \; (3a)$

Radius of equation $(2) = r = \sqrt{g^2 + f^2 - c}$

i.e. $\;$ $r = \sqrt{\left(\dfrac{2 \lambda - 3}{2 - 2 \lambda}\right)^2 + \left(\dfrac{6 \lambda - 9}{2 - 2 \lambda}\right)^2 - \left(\dfrac{45 - 30 \lambda}{1 - \lambda}\right)}$ $\;\;\; \cdots \; (3b)$

The limiting points are the centers of circles of the system whose radii are zero.

$\therefore \;$ We have from equation $(3b)$

$\left(\dfrac{2 \lambda - 3}{2 - 2 \lambda}\right)^2 + \left(\dfrac{6 \lambda - 9}{2 - 2 \lambda}\right)^2 - \left(\dfrac{45 - 30 \lambda}{1 - \lambda}\right) = 0$

i.e. $\;$ $\dfrac{4 \lambda^2 - 12 \lambda + 9 + 36 \lambda^2 - 108 \lambda + 81}{4 \left(1 - \lambda\right)^2} = \dfrac{15 \left(3 - 2 \lambda\right)}{1 - \lambda}$

i.e. $\;$ $\dfrac{40 \lambda^2 - 120 \lambda + 90}{4 \left(1 - \lambda\right)} = 15 \left(3 - 2 \lambda\right)$ $\;\;\;$ provided $\;$ $1 - \lambda \neq 0$

i.e. $\;$ $20 \lambda^2 - 60 \lambda + 45 = 30 \left(3 - 2 \lambda\right) \left(1 - \lambda\right)$

i.e. $\;$ $4 \lambda^2 - 12 \lambda + 9 = 6 \left(3 - 5 \lambda + 12 \lambda^2\right)$

i.e. $\;$ $8 \lambda^2 - 18 \lambda + 9 = 0$

$\implies$ $\lambda = \dfrac{3}{4}$ $\;$ or $\;$ $\lambda = \dfrac{3}{2}$

Substituting the values of $\lambda$ in equation $(3a)$ gives the limiting points.

When $\;$ $\lambda = \dfrac{3}{4}$, $\;$ the limiting point is

$\left(\dfrac{3 - 2 \times \dfrac{3}{4}}{2 - 2 \times\dfrac{3}{4}}, \; \dfrac{9 - 6 \times \dfrac{3}{4}}{2 - 2 \times \dfrac{3}{4}}\right) = \left(3, 9\right)$

When $\;$ $\lambda = \dfrac{3}{2}$, $\;$ the limiting point is

$\left(\dfrac{3 - 2 \times \dfrac{3}{2}}{2 - 2 \times \dfrac{3}{2}}, \; \dfrac{9 - 6 \times \dfrac{3}{2}}{2 - 2 \times \dfrac{3}{2}}\right) = \left(0, 0\right)$