# 1.2 CONICS

Conic: A conic is defined as the locus of a point which moves such that its distance from a fixed point is always ‘e’

times its distance from a fixed straight line.

Focus: The fixed point is called the focus of the conic.

Directrix: The fixed straight line is called the directrix of the conic.

Eccentricity: The constant ratio is called the eccentricity of the conic.

General equation of a conic  ax2  +  2hxy  + by2 + 2gx + 2fy + c = 0  represents

(i) a circle if a = b and h = 0.

(ii) a parabola if  h2 = ab.

(iii) an ellipse if h2 < ab.

(iv) a hyperbola if h2> ab.

$\color {purple} {Example\ 1\ .}\ \color {red} {Prove\ that}\ the\ equation\ x^2\ +\ 6\ x\ y\ +\ 9y^2\ +\ 4\ x\ +\ 12\ y\ -\ 5\ =\ 0\ \hspace{10cm}$$\color {red} {is\ a\ parobala}\ \hspace{5cm}$
$\color {blue} {Soln:}\ x^2\ +\ 6\ x\ y\ +\ 9y^2\ +\ 4\ x\ +\ 12\ y\ -\ 5\ =\ 0\ ———- (1)\ \hspace{10cm}$
$\hspace{2cm}\ Condition\ for\ (1)\ to\ represent\ parabola\ is\ h^2\ =\ ab\ \hspace{8cm}$
$Comparing\ with\ a\ x^2\ +\ 2h\ x\ y\ +\ by^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ = 0$
$We\ get\ a\ =\ 1,\ b\ =\ 9\ \hspace{5cm}\ 2h\ =\ 6,\ \implies\ h\ =\ 3\ \hspace{5cm}$
$h^2\ =\ ab$
$\therefore\ (1)\ represents\ a\ parabola\ \hspace{10cm}$

$\color {purple} {Example\ 2\ .}\ \color {red} {Prove\ that}\ the\ equation\ x^2\ +\ 6\ x\ y\ +\ 9y^2\ +\ 4\ x\ +\ 12\ y\ -\ 5\ =\ 0\ \hspace{10cm}$$\color {red} {is\ a\ parobala}\ \hspace{5cm}$
$\color {blue} {Soln:}\ x^2\ +\ 6\ x\ y\ +\ 9y^2\ +\ 4\ x\ +\ 12\ y\ -\ 5\ =\ 0\ ———- (1)\ \hspace{10cm}$
$\hspace{2cm}\ Condition\ for\ (1)\ to\ represent\ parabola\ is\ h^2\ =\ ab\ \hspace{8cm}$
$Comparing\ with\ a\ x^2\ +\ 2h\ x\ y\ +\ by^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ = 0$
$We\ get\ a\ =\ 1,\ b\ =\ 9\ \hspace{5cm}\ 2h\ =\ 6,\ \implies\ h\ =\ 3\ \hspace{5cm}$
$h^2\ =\ ab$
$3^2\ =\ 1(9)$
$9\ =\ 9$
$\therefore\ (1)\ represents\ a\ parabola\ \hspace{10cm}$

$\color {purple} {Example\ 3\ .}\ \color {red} {Show\ that}\ the\ equation\ x^2\ +\ 4y^2\ -\ 4\ x\ +\ 24\ y\ +\ 31\ =\ 0\ \hspace{10cm}$$\color {red} {represents\ an\ ellipse}\ \hspace{5cm}$
$\color {blue} {Soln:}\ x^2\ +\ 4y^2\ -\ 4\ x\ +\ 24\ y\ +\ 31\ =\ 0\ ———- (1)\ \hspace{10cm}$
$\hspace{2cm}\ Condition\ for\ (1)\ to\ represent\ an\ ellipse\ is\ h^2\ -\ ab\ \lt\ 0\ \hspace{8cm}$
$Comparing\ with\ a\ x^2\ +\ 2h\ x\ y\ +\ by^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ = 0$
$We\ get\ a\ =\ 1,\ b\ =\ 4\ \hspace{5cm}\ 2h\ =\ 0,\ \implies\ h\ =\ 0\ \hspace{5cm}$
$\hspace{2cm}\ h^2\ -\ ab\ =\ (0)^2\ -\ 1(4)\ =\ -\ 4\ \lt\ 0\ \hspace{8cm}$
$\therefore\ (1)\ represents\ an\ ellipse\ \hspace{10cm}$

$\color {purple} {Example\ 4\ .}\ \color {red} {Check}\ whether\ the\ conic\ 2\ x^2\ -\ 16\ x\ y\ +\ 8\ y^2\ -\ y\ +\ 3\ = 0\ \color {red} {represent\ a\ hyperbola}\ \hspace{5cm}$
$\color {blue} {Soln:}\ 2\ x^2\ -\ 16\ x\ y\ +\ 8\ y^2\ -\ y\ +\ 3\ ———- (1)\ \hspace{10cm}$
$\hspace{2cm}\ Condition\ for\ (1)\ to\ represent\ hyperbola\ is\ h^2\ -\ ab\ \gt\ 0\ \hspace{8cm}$
$Comparing\ with\ a\ x^2\ +\ 2h\ x\ y\ +\ by^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ = 0$
$We\ get\ a\ =\ 2,\ b\ =\ 8\ \hspace{5cm}\ 2h\ =\ – 16,\ \implies\ h\ =\ -\ 8\ \hspace{5cm}$
$\hspace{2cm}\ h^2\ -\ ab\ =\ (-8)^2\ -\ 2(4)\ =\ 64\ -\ 8\ =\ 56\ \gt\ 0\ \hspace{8cm}$
$\therefore\ (1)\ represents\ a\ hyperbola\ \hspace{10cm}$

### General Equation of a Conic

‘S’  denotes Focus

Line XM denotes Directrix

SP / PM  = e

Note:

(i) If  e< 1,  the conic is called an ellipse.

(ii) If  e = 1,  the conic is called aparabola.

(iii) If e>1, the conic is called a hyperbola.

### Equation of a Conic with its focus at (x1 , y1 ) and the directrix ax + by + c = 0

$Let\ the\ focus\ be\ S(x_1,\ y_1)\ and\ directrix\ be\ the\ line\ a\ x\ +\ b\ y\ +\ c\ =\ 0\ \hspace{10cm}$
$P(x,\ y)\ be\ any\ point\ on\ it\ \hspace{18cm}$
$SP\ =\ \sqrt{(x\ -\ x_1)^2\ +\ (y\ -\ y_1)^2}\ \hspace{18cm}$
$PM\ =\ Perpendicular\ distance\ of\ P\ from\ the\ line\ a\ x\ +\ b\ y\ +\ c\ =\ 0\ \hspace{10cm}$
$=\ \pm\ \frac{a\ x\ +\ b\ y\ +\ c}{\sqrt{a^2\ +\ b^2}}\ \hspace{15cm}$
$\color {purple} {Example\ 5\ .}\ \color {red} {Find\ the\ equation\ of\ the\ parabola}\ with\ focus\ at\ (1,\ -1)\ \hspace{10cm}$$and\ directrix\ x\ -\ y\ =\ 0.\ \hspace{5cm}$
$\color {blue} {Soln:}\ For\ parabola\ e\ =\ 1\ \hspace{15cm}$
$\hspace{2cm}\ Given\ Focus\ is\ S(1,\ -\ 1)\ and\ directrix\ is\ x\ -\ y\ =\ 0.\ \hspace{8cm}$
$Always\ \frac{SP}{PM}\ =\ e\ =\ 1\ \hspace{10cm}$
$\frac{\sqrt{(x\ -\ x_1)^2\ +\ (y\ -\ y_1)^2}}{\pm\ \frac{a\ x\ +\ b\ y\ +\ c}{\sqrt{a^2\ +\ b^2}}}\ =\ 1\ \hspace{10cm}$
$\frac{\sqrt{(x\ -\ 1)^2\ +\ (y\ +\ 1)^2}}{\pm\ \frac{x\ -\ y}{\sqrt{(1)^2\ +\ (-1)^2}}}\ =\ 1\ \hspace{10cm}$
$\sqrt{(x\ -\ 1)^2\ +\ (y\ +\ 1)^2}\ =\ \pm\ \frac{x\ -\ y}{\sqrt{2}}\ \hspace{10cm}$
$(x\ -\ 1)^2\ +\ (y\ +\ 1)^2\ =\ \frac{(x\ -\ y)^2}{2}\ \hspace{10cm}$
$\hspace{2cm}\ 2(x^2\ -\ 2\ x\ +\ 1\ +\ y^2\ +\ 2\ y\ +\ 1)\ =\ x^2\ +\ y^2\ -\ 2\ x\ y\ \hspace{8cm}$
$\hspace{2cm}\ 2\ x^2\ -\ 4\ x\ +\ 2\ +\ 2\ y^2\ +\ 4\ y\ +\ 2\ -\ x^2\ -\ y^2\ +\ 2\ x\ y\ =\ 0\ \hspace{10cm}$
$\hspace{2cm}\ 2\ x^2\ -\ 4\ x\ +\ 2\ +\ 2\ y^2\ +\ 4\ y\ +\ 2\ -\ x^2\ -\ y^2\ +\ 2\ x\ y\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ \boxed {x^2\ +\ 2\ x\ y\ -\ 4\ x\ +\ y^2\ +\ 4\ y\ +\ 4\ =\ 0}\ \hspace{8cm}$

$\color {purple} {Example\ 6\ .}\ \color {red} {Find\ the\ equation\ of\ the\ parabola}\ with\ focus\ at\ (2,\ 1)\ \hspace{10cm}$$and\ directrix\ 2x\ +\ y\ +\ 1\ =\ 0.\ \hspace{5cm}$
$\color {blue} {Soln:}\ For\ parabola\ e\ =\ 1\ \hspace{15cm}$
$\hspace{2cm}\ Given\ Focus\ is\ S(2,\ 1)\ and\ directrix\ is\ 2\ x\ +\ y\ +\ 1\ =\ 0.\ \hspace{8cm}$
$Always\ \frac{SP}{PM}\ =\ e\ =\ 1\ \hspace{10cm}$
$\frac{\sqrt{(x\ -\ x_1)^2\ +\ (y\ -\ y_1)^2}}{\pm\ \frac{a\ x\ +\ b\ y\ +\ c}{\sqrt{a^2\ +\ b^2}}}\ =\ 1\ \hspace{10cm}$
$\frac{\sqrt{(x\ -\ 2)^2\ +\ (y\ -\ 1)^2}}{\pm\ \frac{2x\ +\ y\ +\ 1}{\sqrt{(2)^2\ +\ (1)^2}}}\ =\ 1\ \hspace{10cm}$
$\sqrt{(x\ -\ 2)^2\ +\ (y\ -\ 1)^2}\ =\ \pm\ \frac{2x\ +\ y\ +\ 1}{\sqrt{5}}\ \hspace{10cm}$
$(x\ -\ 2)^2\ +\ (y\ -\ 1)^2\ =\ \frac{(2x\ +\ y\ +\ 1)^2}{5}\ \hspace{10cm}$
$\hspace{2cm}\ 5(x^2\ -\ 4\ x\ +\ 4\ +\ y^2\ -\ 2\ y\ +\ 1)\ =\ 4x^2\ +\ y^2\ +\ 1\ +\ 4\ x\ y\ + 4\ x\ +\ 2y\ \hspace{8cm}$
$\hspace{2cm}\ 5\ x^2\ -\ 20\ x\ +\ 20\ +\ 5\ y^2\ -\ 10\ y\ +\ 5\ -\ 4\ x^2\ -\ y^2\ -\ 1\ -\ 4\ x\ y\ -\ 4x\ -\ 2y\ =\ 0\ \hspace{10cm}$
$\hspace{2cm}\ x^2\ +\ 4\ y^2\ -\ 4\ x\ y\ -\ 24\ x\ -\ 12\ y\ +\ 24\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ \boxed {x^2\ +\ 4\ y^2\ -\ 4\ x\ y\ -\ 24\ x\ -\ 12\ y\ +\ 24\ =\ 0}\ \hspace{8cm}$

$\color {purple} {Example\ 7\ .}\ \color {red} {Find\ the\ equation\ of\ the\ Ellipse}\ with\ focus\ (2,\ 3)\ and\ directrix\ x\ =\ 7\ and\ e\ =\ \frac{1}{2}\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\hspace{2cm}\ Given\ Focus\ is\ S(2,\ 3)\ and\ directrix\ is\ x\ -\ 7\ =\ 0,\ e\ =\ \frac{1}{2}\ \hspace{8cm}$
$\frac{\sqrt{(x\ -\ x_1)^2\ +\ (y\ -\ y_1)^2}}{\pm\ \frac{a\ x\ +\ b\ y\ +\ c}{\sqrt{a^2\ +\ b^2}}}\ =\ e\ \hspace{10cm}$
$\frac{\sqrt{(x\ -\ 2)^2\ +\ (y\ -\ 3)^2}}{\pm\ \frac{x\ -\ 7}{\sqrt{(1)^2\ +\ (0)^2}}}\ =\ \frac{1}{2}\ \hspace{10cm}$
$2\ \sqrt{(x\ -\ 2)^2\ +\ (y\ -\ 3)^2}\ =\ \pm\ \frac{x\ -\ 7}{\sqrt{1}}\ \hspace{10cm}$
$4\ (x\ -\ 2)^2\ +\ (y\ -\ 3)^2\ =\ \frac{(x\ -\ 7)^2}{1}\ \hspace{10cm}$
$\hspace{2cm}\ 4(x^2\ -\ 4\ x\ +\ 4\ +\ y^2\ -\ 6\ y\ +\ 9)\ =\ x^2\ -\ 14\ x\ +\ 49\ \hspace{8cm}$
$\hspace{2cm}\ 4\ x^2\ -\ 16\ x\ +\ 16\ -\ 24\ y\ +\ 36\ -\ x^2\ +\ 14\ x\ -\ 49\ =\ 0\ \hspace{10cm}$
$\hspace{2cm}\ 3\ x^2\ -\ 2\ x\ +\ 4\ y^2\ -\ 24\ y\ +\ 3\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ 3\ x^2\ +\ 4\ y^2\ – 2\ x\ -\ 24\ y\ +\ 3\ =\ 0\ \hspace{8cm}$

$Condition\ for\ General\ equation\ of\ a\ conic\ a\ x^2\ +\ 2\ h\ x\ y\ +\ b\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{7cm}$$to\ represent\ pair\ of\ straight\ lines\ is\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ =\ 0\ \hspace{5cm}$
$\color {purple} {Example\ 8\ .}\ \color {red} {Prove\ that}\ equation\ 6\ x^2\ +\ 13\ x\ y\ +\ 6\ y^2\ +\ 8\ x\ +\ 7\ y\ +\ 2\ =\ 0\ \hspace{7cm}$$\color {red} {represents\ a\ pair\ of\ straight\ lines}\ \hspace{5cm}$
$\color {blue} {Soln:}\ Given\ 6\ x^2\ +\ 13\ x\ y\ +\ 6\ y^2\ +\ 8\ x\ +\ 7\ y\ +\ 2\ =\ 0\ ———-(1)\ \hspace{8cm}$
$\hspace{2cm}\ a\ x^2\ +\ 2\ h\ x\ y\ +\ b\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}$
$\hspace{2cm}\ comparing\ we\ get\ \hspace{8cm}$
$\hspace{2cm}\ a\ =\ 6\ \hspace{2cm}\ 2\ h\ =\ 13\ \hspace{2cm}\ b\ =\ 6\ \hspace{2cm}\ 2\ g\ =\ 8\ \hspace{2cm}\ 2\ f\ =\ 7\ \hspace{2cm}\ c\ =\ 2$
$\hspace{6cm}\ h\ =\ \frac{13}{2}\ \hspace{4cm}\ g\ =\ 4\ \hspace{4cm}\ f\ =\ \frac{7}{2}\ \hspace{4cm}$
$\hspace{2cm}\ To\ claim\ equation\ (1)\ represents\ a\ pair\ of\ straight\ lines\ \hspace{8cm}$
$\hspace{2cm}\ \therefore\ i.e\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ L.H.S\ =\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ \hspace{8cm}$
$\hspace{2cm}\ =\ 6\ (6)\ (2)\ +\ 2\ (\frac{7}{2})\ (4)\ (\frac{13}{2})\ -\ 6\ (\frac{7}{2}) ^2\ -\ 6(4)^2\ -\ 2\ (\frac{13}{2}) ^2\ \hspace{8cm}$
$\hspace{2cm}\ =\ 72\ +\ \frac{364}{2}\ -\ 6\ (\frac{49}{4})\ -\ 6\ (16)\ -\ 2\ (\frac{169}{4})\ \hspace{10cm}$
$=\ 72\ +\ 182\ -\ \frac{294}{4}\ -\ 96\ -\ \frac{338}{4}\ \hspace{10cm}$
$=\ 158\ -\ (\frac{294\ +\ 338}{4})\ \hspace{10cm}$
$=\ 158\ -\ \frac{632}{4}\ \hspace{10cm}$
$=\ 158\ -\ 158\ \hspace{10cm}$
$=\ 0\ =\ R.\ H.\ S\ \hspace{10cm}$
$\hspace{2cm}\ \therefore\ the\ given\ equation\ (1)\ represents\ a\ pair\ of\ straighlt\ lines\ \hspace{8cm}$

$\color {purple} {Example\ 9\ .}\ \color {red} {Show\ that\ the\ second\ degree\ equation}\ \hspace{5cm}$$12\ x^2\ +\ 7\ x\ y\ -\ 10\ y^2\ +\ 13\ x\ +\ 45\ y\ -\ 35\ =\ 0\ \hspace{7cm}$$\color{red}{in\ x\ and\ y\ represents\ a\ pair\ of\ straight\ lines}\ \hspace{5cm}$
$\color {blue} {Soln:}\ Given\ 12\ x^2\ +\ 7\ x\ y\ -\ 10\ y^2\ +\ 13\ x\ +\ 45\ y\ -\ 35\ =\ 0\ ———-(1)\ \hspace{8cm}$
$\hspace{2cm}\ a\ x^2\ +\ 2\ h\ x\ y\ +\ b\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ comparing\ we\ get\ \hspace{8cm}$
$\hspace{2cm}\ a\ =\ 12\ \hspace{2cm}\ 2\ h\ =\ 7\ \hspace{2cm}\ b\ =\ -\ 10\ \hspace{2cm}\ 2\ g\ =\ 13\ \hspace{2cm}\ 2\ f\ =\ \ 45\ \hspace{2cm}\ c\ =\ -\ 35$
$\hspace{6cm}\ h\ =\ \frac{7}{2}\ \hspace{4cm}\ g\ =\ \frac{13}{2}\ \hspace{4cm}\ f\ =\ \frac{45}{2}\ \hspace{4cm}$
$\hspace{2cm}\ To\ claim\ equation\ (1)\ represents\ a\ pair\ of\ straight\ lines\ \hspace{8cm}$
$\hspace{2cm}\ \therefore\ i.e\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ L.H.S\ =\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ \hspace{8cm}$
$\hspace{2cm}\ =\ 12\ (-10)\ (-35)\ +\ 2\ (\frac{45}{2})\ (\frac{13}{2})\ (\frac{7}{2})\ -\ 12\ (\frac{45}{2}) ^2\ +\ 10\ (\frac{13}{2}) ^2\ +\ 35\ (\frac{7}{2}) ^2\ \hspace{8cm}$
$\hspace{2cm}\ =\ 4200\ +\ \frac{4095}{4}\ -\ 12\ (\frac{2025}{4})\ +\ 10\ (\frac{169}{4})\ +\ 35\ (\frac{49}{4})\ \hspace{10cm}$
$\hspace{2cm}\ =\ 4200\ +\ \frac{4095}{4}\ -\ 3(2025)\ +\ 5(\frac{169}{2})\ +\ \frac{1715}{4}\ \hspace{10cm}$
$\hspace{2cm}\ =\ 4200\ +\ \frac{4095}{4}\ -\ 6075\ +\ \frac{845}{2}\ +\ \frac{1715}{4}\ \hspace{10cm}$
$=\ -\ 1875\ +\ (\frac{4095\ +\ 1690\ +\ 1715}{4})\ \hspace{10cm}$
$=\ -\ 1875\ +\ (\frac{7500}{4})\ \hspace{10cm}$
$=\ -\ 1875\ +\ 1875\ \hspace{10cm}$
$=\ 0\ =\ R.\ H.\ S\ \hspace{10cm}$
$\hspace{2cm}\ \therefore\ the\ given\ equation\ (1)\ represents\ a\ pair\ of\ straighlt\ lines\ \hspace{8cm}$

$\color {purple} {Example\ 10\ .}\ \color {red} {Find\ c\ if}\ \ 2\ x^2\ +\ 3\ x\ y\ -\ 2\ y^2\ -\ 5\ x\ +\ 5\ y\ +\ c\ =\ 0\ \hspace{7cm}$$\color {red} {represents\ a\ pair\ of\ straight\ lines}\ \hspace{5cm}$
$\color {blue} {Soln:}\ Given\ 2\ x^2\ +\ 3\ x\ y\ -\ 2\ y^2\ -\ 5\ x\ +\ 5\ y\ +\ c\ =\ 0\ ———-(1)\ \hspace{8cm}$
$\hspace{2cm}\ a\ x^2\ +\ 2\ h\ x\ y\ +\ b\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{8cm}$
$comparing\ we\ get\ \hspace{10cm}$
$\hspace{1cm}\ a\ =\ 2\ \hspace{1cm}\ 2\ h\ =\ 3\ \hspace{1cm}\ b\ =\ -\ 2\ \hspace{1cm}\ 2\ g\ =\ -\ 5\ \hspace{1cm}\ 2\ f\ =\ 5\ \hspace{1cm}\ c\ =\ c$
$\hspace{5cm}\ h\ =\ \frac{3}{2}\ \hspace{5cm}\ g\ =\ -\ \frac{5}{2}\ \hspace{3cm}\ f\ =\ \frac{5}{2}\ \hspace{3cm}$
$\hspace{2cm}\ Given\ that\ \ (1)\ represents\ a\ pair\ of\ straight\ lines\ \hspace{8cm}$
$\hspace{2cm}\ i.e\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ 2\ (-2)\ (c)\ +\ 2\ (\frac{5}{2})\ (-\frac{5}{2})\ (\frac{3}{2})\ -\ 2\ (\frac{5}{2}) ^2\ +\ 2(-\frac{5}{2})^2\ -\ c\ (\frac{3}{2}) ^2\ =\ 0\ \hspace{10cm}$
$\hspace{2cm}\ -\ 4\ c\ -\ \frac{75}{4}\ -\ 2\ (\frac{25}{4})\ +\ 2\ (\frac{25}{4})\ -\ c\ (\frac{9}{4})\ =\ 0\ \hspace{10cm}$
$\hspace{2cm}\ \frac{-\ 16\ c\ -\ 75\ -\ 50\ +\ 50\ -\ 9\ c}{4}\ =\ 0\ \hspace{10cm}$
$-\ 25\ c\ -\ 75\ =\ 0\ \hspace{10cm}$
$25\ c\ =\ -\ 75\ \hspace{10cm}$
$\boxed {c\ =\ -\ 3}\ \hspace{10cm}$

$\color {purple} {Example\ 11:}\ \color {red} {Find\ the\ value\ of\ α}\ if\ \ α\ x^2\ -\ 10\ x\ y\ + \ 5\ x\ +\ 12\ y^2\ -\ 16y\ -\ 3\ =\ 0\ \hspace{7cm}$$\color {red} {represents\ a\ pair\ of\ straight\ lines}\ \hspace{5cm}$
$\color {blue} {Soln:}\ Given\ α\ x^2\ -\ 10\ x\ y\ + \ 5\ x\ +\ 12\ y^2\ -\ 16y\ -\ 3\ =\ 0\ ———-(1)\ \hspace{8cm}$
$\hspace{2cm}\ a\ x^2\ +\ 2\ h\ x\ y\ +\ b\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}$
$\hspace{2cm}\ comparing\ we\ get\ \hspace{8cm}$
$\hspace{1cm}\ a\ =\ α\ \hspace{1cm}\ 2\ h\ =\ -\ 10\ \hspace{1cm}\ b\ =\ 12\ \hspace{1cm}\ 2\ g\ =\ 5\ \hspace{1cm}\ 2\ f\ =\ – 16\ \hspace{1cm}\ c\ =\ – 3$
$\hspace{5cm}\ h\ =\ -\ 5\ \hspace{3cm}\ g\ =\ \frac{5}{2}\ \hspace{3cm}\ f\ =\ -\ 8\ \hspace{3cm}$
$\hspace{2cm}\ Given\ that\ \ (1)\ represents\ a\ pair\ of\ straight\ lines\ \hspace{8cm}$
$\hspace{2cm}\ i.e\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ α\ (12)\ (-3)\ +\ 2\ (-8)\ (\frac{5}{2})\ (-5)\ -\ α\ (-8) ^2\ -\ 12(\frac{5}{2})^2\ +\ 3\ (-5) ^2\ =\ 0\ \hspace{10cm}$
$\hspace{2cm}\ -\ 36\ α\ +\ 200\ -\ 64\ α\ -\ 12\ (\frac{25}{4})\ +\ 3(25)\ =\ 0\ \hspace{10cm}$
$\hspace{2cm}\ -\ 100\ α\ +\ 200\ -\ 75\ +\ 75\ =\ 0\ \hspace{10cm}$
$-\ 100\ α\ +\ 200\ =\ 0\ \hspace{10cm}$
$-100\ α\ =\ -\ 200\ \hspace{10cm}$
$\boxed {α\ =\ 2}\ \hspace{10cm}$

### Exercise Problems

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1\ .}\ \color {red} {Define}\ Conic\ \hspace{18cm}$
$\color {purple} {2\ .}\ \color {red} {Prove\ that}\ the\ equation\ x^2\ -\ 2\ x\ y\ +\ y^2\ -\ 16\ x\ -\ 12\ y\ +\ 22\ = 0\ is\ a\ parabola\ \hspace{5cm}$
$\color {purple} {3\ .}\ \color {red} {Show\ that}\ the\ equation\ 7x^2\ +\ 3xy\ +\ 2y^2\ -\ \ x\ +\ 2\ y\ -\ 1\ =\ 0\ \hspace{10cm}$$\color {red} {represents\ an\ ellipse}\ \hspace{5cm}$
$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {4\ .}\ \color {red} {Show\ that}\ the\ equation\ 4\ x^2\ +\; 10\ x\ y\ +\ y^2\ -\ 2\ x\ +\ 5\ y\ -\ 3\ = 0\ \color {red} {represents\ a\ hyperbola}\ \hspace{5cm}$
$\color {purple} {5\ .}\ \color {red} {Find\ the\ equation\ of\ the\ parabola}\ with\ focus\ at\ (-1,\ -2)\ and\ x\ +\ 2\ y\ =\ 0\ as\ its\ directrix\ \hspace{5cm}$
$\color {purple} {6\ .}\ \color{red}{What\ is\ the\ condition\ for\ the\ conic}\ a\ x^2\ +\ 2\ h\ x\ y\ +\ b\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{7cm}$$\color {red} {represents\ a\ pair\ of\ straight\ lines}\ \hspace{5cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {7\ .}\ \color{red}{Find\ the\ equation\ of\ the\ Ellipse}\ with\ focus\ (-1,\ -3)\ and\ directrix\ x\ -\ 2y\ =\ 0\ and\ e\ =\ \frac{4}{5}\ \hspace{5cm}$
$\color {purple} {8\ .}\ \color{red}{Prove\ that}\ equation\ 3\ x^2\ +\ 7\ x\ y\ +\ 2\ y^2\ +\ 5\ x\ +\ 5\ y\ +\ 2\ =\ 0\ \hspace{7cm}$$\color {red} {represents\ a\ pair\ of\ straight\ lines}\ \hspace{5cm}$
$\color {purple} {9\ .}\ For\ the\ quadratic\ equation\ 2\ x^2\ +\ 7\ x\ y\ +\ 3\ y^2\ +\ 13\ x\ -\ \ y\ -\ 24\ =\ 0\ \hspace{7cm}$$\color {red} {identify\ the\ conic}\ \hspace{5cm}$