1.2 – N – CONICS – Exercise Problems with solutions

\[\underline{PART\ -\ A}\]
\[1.\ Define\ Conic\ \hspace{18cm}\]
\[Soln:\ A\ conic\ is\ defined\ as\ the\ locus\ of\ a\ point\ which\ moves\ such\ that\ its\ distance\ from\ a\ fixed\ point\ is\ \hspace{5cm}\]\[always ‘e’\ times\ its\ distance\ from\ a\ fixed\ straight line.\ \hspace{10cm}\]
\[2.\ Prove\ that\ the\ equation\ x^2\ -\ 2\ x\ y\ +\ y^2\ -\ 16\ x\ -\ 12\ y\ +\ 22\ = 0\ is\ a\ parabola\ \hspace{5cm}\]

Soln:    x2  –  2xy  + y2 – 16x – 12y + 22 = 0                    —————–   ( 1 )

Condition for  ( 1 ) to represent parabola is  h2 = ab

From ( 1 )   a =  1,    b = 1

2h = -2  ⇒  h = -1

h2 = ab

(-1)2  =   1 ( 1)

1 = 1

1 = 1.                 ∴  ( 1 )   represents a parabola.

3. Show that the equation  7x2  +  3xy  + 2y2 – x + 2y – 1 = 0  represents an ellipse.

Soln:  Given    7x2  +  3xy  + 2y2 – x + 2y – 1 = 0  –––– (1)

ax2  +  2hxy  + by2 + 2gx + 2fy + c = 0    ––––– (2)

Comparing, we get

a = 7                2h = 3              b = 2

h = 3/2

h2 –  ab =  (3/2)–  7(2)  =  9/4 – 14 =  (9 -56)/14 = -47/14 < 0

Given equation ( 1 ) represents ellipse.

New 3D Puzzle
\[\underline{PART\ -\ B}\]
\[4.\ Show\ that\ the\ equation\ 4\ x^2\ +\; 10\ x\ y\ +\ y^2\ -\ 2\ x\ +\ 5\ y\ -\ 3\ = 0\ represents\ a\ hyperbola\ \hspace{5cm}\]
\[Soln:\ 4\ x^2\ +\; 10\ x\ y\ +\ y^2\ -\ 2\ x\ +\ 5\ y\ -\ 3\ = 0\ ———- (1)\ \hspace{10cm}\]
\[\hspace{2cm}\ Condition\ for\ (1)\ to\ represent\ hyperbola\ is\ h^2\ -\ ab\ \gt\ 0\ \hspace{8cm}\]
\[From\ (1)\ a\ =\ 4,\ b\ =\ 1\ \hspace{10cm}\]
\[2h\ =\ 10,\ \implies\ h\ =\ 5\ \hspace{10cm}\]
\[\hspace{2cm}\ h^2\ -\ ab\ =\ (5)^2\ -\ 4(1)\ =\ 25\ -\ 1\ =\ 24\ \gt\ 0\ \hspace{8cm}\]
\[\therefore\ (1)\ represents\ a\ hyperbola\ \hspace{10cm}\]
\[5.\ Find\ the\ equation\ of\ the\ parabola\ with\ focus\ at\ (-1,\ -2)\ and\ x\ +\ 2\ y\ =\ 0\ as\ its\ directrix\ \hspace{5cm}\]
\[Soln:\ For\ parabola\ e\ =\ 1\ \hspace{15cm}\]
\[\hspace{2cm}\ Given\ Focus\ is\ S(-1,\ -2)\ and\ directrix\ is\ x\ +\ 2\ y\ =\ 0.\ \hspace{8cm}\]
\[Always\ \frac{SP}{PM}\ =\ e\ =\ 1\ \hspace{10cm}\]
\[\frac{\sqrt{(x\ -\ x_1)^2\ +\ (y\ -\ y_1)^2}}{\pm\ \frac{a\ x\ +\ b\ y\ +\ c}{\sqrt{a^2\ +\ b^2}}}\ =\ 1\ \hspace{10cm}\]
\[\frac{\sqrt{(x\ +\ 1)^2\ +\ (y\ +\ 2)^2}}{\pm\ \frac{x\ +\ 2\ y}{\sqrt{(1)^2\ +\ (2)^2}}}\ =\ 1\ \hspace{10cm}\]
\[\sqrt{(x\ +\ 1)^2\ +\ (y\ +\ 2)^2}\ =\ \pm\ \frac{x\ +\ 2\ y}{\sqrt{5}}\ \hspace{10cm}\]
\[(x\ +\ 1)^2\ +\ (y\ +\ 2)^2\ =\ \frac{(x\ +\ 2\ y)^2}{5}\ \hspace{10cm}\]
\[\hspace{2cm}\ 5(x^2\ +\ 2\ x\ +\ 1\ +\ y^2\ +\ 4\ y\ +\ 4)\ =\ x^2\ +\ 4\ y^2\ +\ 4\ x\ y\ \hspace{8cm}\]
\[\hspace{2cm}\ 5\ x^2\ +\ 10\ x\ +\ 5\ +\ 5\ y^2\ +\ 20\ y\ +\ 20\ -\ x^2\ -\ 4\ y^2\ -\ 4\ x\ y\ =\ 0\ \hspace{10cm}\]
\[\hspace{2cm}\ 4\ x^2\ -\ 4\ x\ y\ +\ 10\ x\ +\ y^2\ -\ 20\ y\ +\ 25\ =\ 0\ \hspace{8cm}\]
\[\hspace{2cm}\ \boxed {4\ x^2\ -\ 4\ x\ y\ +\ 10\ x\ +\ y^2\ -\ 20\ y\ +\ 25\ =\ 0}\ \hspace{8cm}\]

\[\underline{PART\ -\ C}\]
\[6.\ Prove\ that\ equation\ 3\ x^2\ +\ 7\ x\ y\ +\ 2\ y^2\ +\ 5\ x\ +\ 5\ y\ +\ 2\ =\ 0\ \hspace{7cm}\]\[represents\ a\ pair\ of\ straight\ lines\ \hspace{5cm}\]
\[Soln:\ Given\ 3\ x^2\ +\ 7\ x\ y\ +\ 2\ y^2\ +\ 5\ x\ +\ 5\ y\ +\ 2\ =\ 0\ ———-(1)\ \hspace{8cm}\]
\[\hspace{2cm}\ a\ x^2\ +\ 2\ h\ x\ y\ +\ b\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{8cm}\]
\[\hspace{2cm}\ comparing\ we\ get\ \hspace{8cm}\]
\[\hspace{2cm}\ a\ =\ 3\ \hspace{2cm}\ 2\ h\ =\ 7\ \hspace{2cm}\ b\ =\ 2\ \hspace{2cm}\ 2\ g\ =\ 5\ \hspace{2cm}\ 2\ f\ =\ \ 5\ \hspace{2cm}\ c\ =\ 2\]
\[\hspace{6cm}\ h\ =\ \frac{7}{2}\ \hspace{4cm}\ g\ =\ \frac{5}{2}\ \hspace{4cm}\ f\ =\ \frac{5}{2}\ \hspace{4cm}\]
\[\hspace{2cm}\ To\ claim\ equation\ (1)\ represents\ a\ pair\ of\ straight\ lines\ \hspace{8cm}\]
\[\hspace{2cm}\ \therefore\ i.e\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ =\ 0\ \hspace{8cm}\]
\[\hspace{2cm}\ L.H.S\ =\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ \hspace{8cm}\]
\[\hspace{2cm}\ =\ 3\ (2)\ (2)\ +\ 2\ (\frac{5}{2})\ (\frac{5}{2})\ (\frac{7}{2})\ -\ 3\ (\frac{-5}{2}) ^2\ -\ 2\ (\frac{5}{2}) ^2\ -\ 2\ (\frac{7}{2}) ^2\ \hspace{8cm}\]
\[\hspace{2cm}\ =\ 12\ +\ \frac{175}{4}\ -\ 3\ (\frac{25}{4})\ -\ 2\ (\frac{25}{4})\ -\ 2\ (\frac{49}{4})\ \hspace{10cm}\]
\[ =\ 12\ +\ \frac{175}{4}\ -\ \frac{75}{4}\ -\ \frac{50}{4}\ -\ \frac{98}{4}\ \hspace{10cm}\]
\[ =\ 12\ +\ (\frac{175\ -\ 75\ -\ 50\ -\ 98}{4})\ \hspace{10cm}\]
\[ =\ 12\ +\ (\frac{175\ -\ 223}{4})\ \hspace{10cm}\]
\[ =\ 12\ +\ \frac{- 48}{4}\ \hspace{10cm}\]
\[ =\ 12\ -\ 12\ \hspace{10cm}\]
\[ =\ 0\ =\ R.\ H.\ S\ \hspace{10cm}\]
\[\hspace{2cm}\ \therefore\ the\ given\ equation\ (1)\ represents\ a\ pair\ of\ straighlt\ lines\ \hspace{8cm}\]
\[7.\ For\ the\ quadratic\ equation\ 2\ x^2\ +\ 7\ x\ y\ +\ 3\ y^2\ +\ 13\ x\ -\ \ y\ -\ 24\ =\ 0\ \hspace{7cm}\]\[identify\ the\ conic\ \hspace{5cm}\]
\[Soln:\ Given\ 2\ x^2\ +\ 7\ x\ y\ +\ 3\ y^2\ +\ 13\ x\ -\ \ y\ -\ 24\ =\ 0\ ———-(1)\ \hspace{8cm}\]
\[\hspace{2cm}\ a\ x^2\ +\ 2\ h\ x\ y\ +\ b\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}\]
\[\hspace{2cm}\ comparing\ we\ get\ \hspace{8cm}\]
\[\hspace{2cm}\ a\ =\ 2\ \hspace{2cm}\ 2\ h\ =\ 7\ \hspace{2cm}\ b\ =\ 3\ \hspace{2cm}\ 2\ g\ =\ 13\ \hspace{2cm}\ 2\ f\ =\ – 1\ \hspace{2cm}\ c\ =\ – 24\]
\[\hspace{6cm}\ h\ =\ \frac{7}{2}\ \hspace{4cm}\ g\ =\ \frac{13}{2}\ \hspace{4cm}\ f\ =\ \frac{-1}{2}\ \hspace{4cm}\]
\[\hspace{2cm}\ To\ claim\ equation\ (1)\ represents\ a\ pair\ of\ straight\ lines\ \hspace{8cm}\]
\[\hspace{2cm}\ \therefore\ i.e\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ =\ 0\ \hspace{8cm}\]
\[\hspace{2cm}\ L.H.S\ =\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ \hspace{8cm}\]
\[\hspace{2cm}\ =\ 2\ (3)\ (-24)\ +\ 2\ (\frac{-1}{2})\ (\frac{13}{2})\ (\frac{7}{2})\ -\ 2\ (\frac{- 1}{2}) ^2\ -\ 3(\frac{13}{2})^2\ +\ 24\ (\frac{7}{2}) ^2\ \hspace{8cm}\]
\[\hspace{2cm}\ =\ -\ 144\ -\ \frac{91}{4}\ -\ 2\ (\frac{1}{4})\ -\ 3\ (\frac{169}{4})\ +\ 24\ (\frac{49}{4})\ \hspace{10cm}\]
\[ =\ -\ 144\ -\ \frac{91}{4}\ -\ \frac{2}{4}\ -\ \frac{507}{4}\ +\ 294\ \hspace{10cm}\]
\[ =\ 150\ -\ (\frac{91\ +\ 2\ +\ 507}{4})\ \hspace{10cm}\]
\[ =\ 150\ -\ \frac{600}{4}\ \hspace{10cm}\]
\[ =\ 150\ -\ 150\ \hspace{10cm}\]
\[ =\ 0\ =\ R.\ H.\ S\ \hspace{10cm}\]
\[\hspace{2cm}\ \therefore\ the\ given\ equation\ (1)\ represents\ a\ pair\ of\ straighlt\ lines\ \hspace{8cm}\]
Dresslily WW

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