1.2 – N – CONICS – Exercise Problems with solutions

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1\ .}\ \color {red} {Define}\ Conic\ \hspace{18cm}$
$\color {blue} {Soln:}\ A\ conic\ is\ defined\ as\ the\ locus\ of\ a\ point\ which\ moves\ such\ that\ its\ distance\ from\ a\ fixed\ point\ is\ \hspace{5cm}$$always ‘e’\ times\ its\ distance\ from\ a\ fixed\ straight line.\ \hspace{10cm}$
$\color {purple} {2\ .}\ \color {red} {Prove\ that}\ the\ equation\ x^2\ -\ 2\ x\ y\ +\ y^2\ -\ 16\ x\ -\ 12\ y\ +\ 22\ = 0\ is\ a\ parabola\ \hspace{5cm}$
$\color {blue} {Soln:}\ x^2\ -\ 2\ x\ y\ +\ y^2\ -\ 16\ x\ -\ 12\ y\ +\ 22\ =\ 0\ ———- (1)\ \hspace{10cm}$
$\hspace{2cm}\ Condition\ for\ (1)\ to\ represent\ parabola\ is\ h^2\ =\ ab\ \hspace{8cm}$
$Comparing\ with\ a\ x^2\ +\ 2h\ x\ y\ +\ by^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ = 0$
$We\ get\ a\ =\ 1,\ b\ =\ 1\ \hspace{5cm}\ 2h\ =\ -\ 2,\ \implies\ h\ =\ -\ 1\ \hspace{5cm}$
$h^2\ =\ ab$
$(-1)^2\ =\ 1(1)$
$1\ =\ 1$
$\therefore\ (1)\ represents\ a\ parabola\ \hspace{10cm}$

$\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}$
$\color {blue} {Soln:}\ 7x^2\ +\ 3xy\ +\ 2y^2\ -\ \ x\ +\ 2\ y\ -\ 1\ =\ 0\ ———- (1)\ \hspace{10cm}$
$\hspace{2cm}\ Condition\ for\ (1)\ to\ represent\ an\ ellipse\ is\ h^2\ -\ ab\ \lt\ 0\ \hspace{8cm}$
$Comparing\ with\ a\ x^2\ +\ 2h\ x\ y\ +\ by^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ = 0$
$We\ get\ a\ =\ 7,\ b\ =\ 2\ \hspace{5cm}\ 2h\ =\ 3,\ \implies\ h\ =\ \frac{3}{2}\ \hspace{5cm}$
$\hspace{2cm}\ h^2\ -\ ab\ =\ (\frac{3}{2})^2\ -\ 7(2)\ =\ \frac{9}{4}\ -\ 14\ =\ -\ \frac{9\ -\ 56}{14}\ =\ -\ \frac{47}{14}\ \lt\ 0\ \hspace{8cm}$
$\therefore\ (1)\ represents\ an\ ellipse\ \hspace{10cm}$

$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {4\ .}\ \color {red} {Show\ that}\ the\ equation\ 4\ x^2\ +\; 10\ x\ y\ +\ y^2\ -\ 2\ x\ +\ 5\ y\ -\ 3\ = 0\ \color {red} {represents\ a\ hyperbola}\ \hspace{5cm}$
$\color {blue} {Soln:}\ 4\ x^2\ +\; 10\ x\ y\ +\ y^2\ -\ 2\ x\ +\ 5\ y\ -\ 3\ = 0\ ———- (1)\ \hspace{10cm}$
$\hspace{2cm}\ Condition\ for\ (1)\ to\ represent\ hyperbola\ is\ h^2\ -\ ab\ \gt\ 0\ \hspace{8cm}$
$Comparing\ with\ a\ x^2\ +\ 2h\ x\ y\ +\ by^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ = 0$
$We\ get\ a\ =\ 4,\ b\ =\ 1\ \hspace{5cm}\ 2h\ =\ 10,\ \implies\ h\ =\ 5\ \hspace{5cm}$
$\hspace{2cm}\ h^2\ -\ ab\ =\ (5)^2\ -\ 4(1)\ =\ 25\ -\ 4\ =\ 21\ \gt\ 0\ \hspace{8cm}$
$\therefore\ (1)\ represents\ a\ hyperbola\ \hspace{10cm}$

$\color {purple} {5\ .}\ \color {red} {Find\ the\ equation\ of\ the\ parabola}\ with\ focus\ at\ (-1,\ -2)\ and\ x\ +\ 2\ y\ =\ 0\ as\ its\ directrix\ \hspace{5cm}$
$\color {blue} {Soln:}\ For\ parabola\ e\ =\ 1\ \hspace{15cm}$
$\hspace{2cm}\ Given\ Focus\ is\ S(-1,\ -2)\ and\ directrix\ is\ x\ +\ 2\ y\ =\ 0.\ \hspace{8cm}$
$Always\ \frac{SP}{PM}\ =\ e\ =\ 1\ \hspace{10cm}$
$\frac{\sqrt{(x\ -\ x_1)^2\ +\ (y\ -\ y_1)^2}}{\pm\ \frac{a\ x\ +\ b\ y\ +\ c}{\sqrt{a^2\ +\ b^2}}}\ =\ 1\ \hspace{10cm}$
$\frac{\sqrt{(x\ +\ 1)^2\ +\ (y\ +\ 2)^2}}{\pm\ \frac{x\ +\ 2\ y}{\sqrt{(1)^2\ +\ (2)^2}}}\ =\ 1\ \hspace{10cm}$
$\sqrt{(x\ +\ 1)^2\ +\ (y\ +\ 2)^2}\ =\ \pm\ \frac{x\ +\ 2\ y}{\sqrt{5}}\ \hspace{10cm}$
$(x\ +\ 1)^2\ +\ (y\ +\ 2)^2\ =\ \frac{(x\ +\ 2\ y)^2}{5}\ \hspace{10cm}$
$\hspace{2cm}\ 5(x^2\ +\ 2\ x\ +\ 1\ +\ y^2\ +\ 4\ y\ +\ 4)\ =\ x^2\ +\ 4\ y^2\ +\ 4\ x\ y\ \hspace{8cm}$
$\hspace{2cm}\ 5\ x^2\ +\ 10\ x\ +\ 5\ +\ 5\ y^2\ +\ 20\ y\ +\ 20\ -\ x^2\ -\ 4\ y^2\ -\ 4\ x\ y\ =\ 0\ \hspace{10cm}$
$\hspace{2cm}\ 4\ x^2\ -\ 4\ x\ y\ +\ 10\ x\ +\ y^2\ +\ 20\ y\ +\ 25\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ \boxed {4\ x^2\ -\ 4\ x\ y\ +\ 10\ x\ +\ y^2\ +\ 20\ y\ +\ 25\ =\ 0}\ \hspace{8cm}$

$\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}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\hspace{2cm}\ The\ condition\ for\ the\ given\ conic\ to\ represent\ a\ pair\ of\ straight\ lines\ is\ \hspace{8cm}$
$\hspace{2cm}\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ =\ 0\ \hspace{8cm}$

$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {7\ .}\ \color{red}{Find\ the\ equation\ of\ the\ Ellipse}\ with\ focus\ (-1,\ -3)\ and\ directrix\ x\ -\ 2y\ =\ 0\ and\ e\ =\ \frac{4}{5}\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\hspace{2cm}\ Given\ Focus\ is\ S(-1,\ -3)\ and\ directrix\ is\ x\ -\ 2y\ =\ 0,\ e\ =\ \frac{4}{5}\ \hspace{8cm}$
$\frac{\sqrt{(x\ -\ x_1)^2\ +\ (y\ -\ y_1)^2}}{\pm\ \frac{a\ x\ +\ b\ y\ +\ c}{\sqrt{a^2\ +\ b^2}}}\ =\ e\ \hspace{10cm}$
$\frac{\sqrt{(x\ +\ 1)^2\ +\ (y\ +\ 3)^2}}{\pm\ \frac{x\ -\ 2y}{\sqrt{(1)^2\ +\ (-2)^2}}}\ =\ \frac{4}{5}\ \hspace{10cm}$
$5\ \sqrt{(x\ +\ 1)^2\ +\ (y\ +\ 3)^2}\ =\ 4\ (\pm\ \frac{x\ -\ 2y}{\sqrt{5}})\ \hspace{10cm}$
$25\ ((x\ +\ 1)^2\ +\ (y\ +\ 3)^2)\ =\ 16\ (\frac{(x\ -\ 2y)^2}{5})\ \hspace{10cm}$
$\hspace{2cm}\ 125(x^2\ +\ 2\ x\ +\ 1\ +\ y^2\ +\ 6\ y\ +\ 9)\ =\ 16(x^2\ -\ 4\ x\ y\ +\ 4\ y^2)\ \hspace{8cm}$
$\hspace{2cm}\ 125\ x^2\ +\ 250\ x\ +\ 125\ +\ 125\ y^2\ +\ 750\ y\ +\ 1125\ -\ 16\ x^2\ +\ 64\ x\ y\ +\ 64\ =\ 0\ \hspace{10cm}$
$\hspace{2cm}\ 109\ x^2\ +\ 64\ x\ y\ +\ 250\ x\ +\ 61\ y^2\ +\ 750\ y\ +\ 1250\ =\ 0\ \hspace{8cm}$

$\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}$
$\color {blue} {Soln:}\ Given\ 3\ x^2\ +\ 7\ x\ y\ +\ 2\ y^2\ +\ 5\ x\ +\ 5\ y\ +\ 2\ =\ 0\ ———-(1)\ \hspace{8cm}$
$\hspace{2cm}\ a\ x^2\ +\ 2\ h\ x\ y\ +\ b\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ comparing\ we\ get\ \hspace{8cm}$
$\hspace{2cm}\ a\ =\ 3\ \hspace{2cm}\ 2\ h\ =\ 7\ \hspace{2cm}\ b\ =\ 2\ \hspace{2cm}\ 2\ g\ =\ 5\ \hspace{2cm}\ 2\ f\ =\ \ 5\ \hspace{2cm}\ c\ =\ 2$
$\hspace{6cm}\ h\ =\ \frac{7}{2}\ \hspace{4cm}\ g\ =\ \frac{5}{2}\ \hspace{4cm}\ f\ =\ \frac{5}{2}\ \hspace{4cm}$
$\hspace{2cm}\ To\ claim\ equation\ (1)\ represents\ a\ pair\ of\ straight\ lines\ \hspace{8cm}$
$\hspace{2cm}\ \therefore\ i.e\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ L.H.S\ =\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ \hspace{8cm}$
$\hspace{2cm}\ =\ 3\ (2)\ (2)\ +\ 2\ (\frac{5}{2})\ (\frac{5}{2})\ (\frac{7}{2})\ -\ 3\ (\frac{-5}{2}) ^2\ -\ 2\ (\frac{5}{2}) ^2\ -\ 2\ (\frac{7}{2}) ^2\ \hspace{8cm}$
$\hspace{2cm}\ =\ 12\ +\ \frac{175}{4}\ -\ 3\ (\frac{25}{4})\ -\ 2\ (\frac{25}{4})\ -\ 2\ (\frac{49}{4})\ \hspace{10cm}$
$=\ 12\ +\ \frac{175}{4}\ -\ \frac{75}{4}\ -\ \frac{50}{4}\ -\ \frac{98}{4}\ \hspace{10cm}$
$=\ 12\ +\ (\frac{175\ -\ 75\ -\ 50\ -\ 98}{4})\ \hspace{10cm}$
$=\ 12\ +\ (\frac{175\ -\ 223}{4})\ \hspace{10cm}$
$=\ 12\ +\ \frac{- 48}{4}\ \hspace{10cm}$
$=\ 12\ -\ 12\ \hspace{10cm}$
$=\ 0\ =\ R.\ H.\ S\ \hspace{10cm}$
$\hspace{2cm}\ \therefore\ the\ given\ equation\ (1)\ represents\ a\ pair\ of\ straighlt\ lines\ \hspace{8cm}$

$\color {purple} {10\ .}\ 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}$
$\color {blue} {Soln:}\ Given\ 2\ x^2\ +\ 7\ x\ y\ +\ 3\ y^2\ +\ 13\ x\ -\ \ y\ -\ 24\ =\ 0\ ———-(1)\ \hspace{8cm}$
$\hspace{2cm}\ a\ x^2\ +\ 2\ h\ x\ y\ +\ b\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}$
$\hspace{2cm}\ comparing\ we\ get\ \hspace{8cm}$
$\hspace{2cm}\ a\ =\ 2\ \hspace{2cm}\ 2\ h\ =\ 7\ \hspace{2cm}\ b\ =\ 3\ \hspace{2cm}\ 2\ g\ =\ 13\ \hspace{2cm}\ 2\ f\ =\ – 1\ \hspace{2cm}\ c\ =\ – 24$
$\hspace{6cm}\ h\ =\ \frac{7}{2}\ \hspace{4cm}\ g\ =\ \frac{13}{2}\ \hspace{4cm}\ f\ =\ \frac{-1}{2}\ \hspace{4cm}$
$\hspace{2cm}\ To\ claim\ equation\ (1)\ represents\ a\ pair\ of\ straight\ lines\ \hspace{8cm}$
$\hspace{2cm}\ \therefore\ i.e\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ L.H.S\ =\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ \hspace{8cm}$
$\hspace{2cm}\ =\ 2\ (3)\ (-24)\ +\ 2\ (\frac{-1}{2})\ (\frac{13}{2})\ (\frac{7}{2})\ -\ 2\ (\frac{- 1}{2}) ^2\ -\ 3(\frac{13}{2})^2\ +\ 24\ (\frac{7}{2}) ^2\ \hspace{8cm}$
$\hspace{2cm}\ =\ -\ 144\ -\ \frac{91}{4}\ -\ 2\ (\frac{1}{4})\ -\ 3\ (\frac{169}{4})\ +\ 24\ (\frac{49}{4})\ \hspace{10cm}$
$=\ -\ 144\ -\ \frac{91}{4}\ -\ \frac{2}{4}\ -\ \frac{507}{4}\ +\ 294\ \hspace{10cm}$
$=\ 150\ -\ (\frac{91\ +\ 2\ +\ 507}{4})\ \hspace{10cm}$
$=\ 150\ -\ \frac{600}{4}\ \hspace{10cm}$
$=\ 150\ -\ 150\ \hspace{10cm}$
$=\ 0\ =\ R.\ H.\ S\ \hspace{10cm}$
$\hspace{2cm}\ \therefore\ the\ given\ equation\ (1)\ represents\ a\ pair\ of\ straighlt\ lines\ \hspace{8cm}$