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\[\color {purple} {1\ .}\ \color {red} {Prove\ that}\ the\ equation\ x^2\ +\ 6\ x\ y\ +\ 9y^2\ +\ 4\ x\ +\ 12\ y\ -\ 5\ =\ 0\ \hspace{10cm}\]\[\color {red} {is\ a\ parobala}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ x^2\ +\ 6\ x\ y\ +\ 9y^2\ +\ 4\ x\ +\ 12\ y\ -\ 5\ =\ 0\ ———- (1)\ \hspace{10cm}\]
\[\hspace{2cm}\ Condition\ for\ (1)\ to\ represent\ parabola\ is\ h^2\ =\ ab\ \hspace{8cm}\]
\[Comparing\ with\ a\ x^2\ +\ 2h\ x\ y\ +\ by^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ = 0\]
\[We\ get\ a\ =\ 1,\ b\ =\ 9\ \hspace{5cm}\ 2h\ =\ 6,\ \implies\ h\ =\ 3\ \hspace{5cm}\]
\[h^2\ =\ ab\]
\[\therefore\ (1)\ represents\ a\ parabola\ \hspace{10cm}\]
\[\color {purple} {2\ .}\ \color {red} {Show\ that}\ the\ equation\ x^2\ +\ 4y^2\ +\ 4\ x\ +\ 24\ y\ +\ 31\ =\ 0\ \hspace{10cm}\]\[\color {red} {represents\ an\ ellipse}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ x^2\ +\ 4y^2\ +\ 4\ x\ +\ 24\ y\ +\ 31\ =\ 0\ ———- (1)\ \hspace{10cm}\]
\[\hspace{2cm}\ Condition\ for\ (1)\ to\ represent\ an\ ellipse\ is\ h^2\ -\ ab\ \lt\ 0\ \hspace{8cm}\]
\[Comparing\ with\ a\ x^2\ +\ 2h\ x\ y\ +\ by^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ = 0\]
\[We\ get\ a\ =\ 1,\ b\ =\ 4\ \hspace{5cm}\ 2h\ =\ 0,\ \implies\ h\ =\ 0\ \hspace{5cm}\]
\[\hspace{2cm}\ h^2\ -\ ab\ =\ (0)^2\ -\ 1(4)\ =\ -\ 4\ \lt\ 0\ \hspace{8cm}\]
\[\therefore\ (1)\ represents\ an\ ellipse\ \hspace{10cm}\]
\[\color {purple} {3\ .}\ \color {red} {Check}\ whether\ the\ conic\ 2\ x^2\ -\ 16\ x\ y\ +\ 8\ y^2\ -\ y\ +\ 3\ = 0\ \color {red} {represent\ a\ hyperbola}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ 2\ x^2\ -\ 16\ x\ y\ +\ 8\ y^2\ -\ y\ +\ 3\ ———- (1)\ \hspace{10cm}\]
\[\hspace{2cm}\ Condition\ for\ (1)\ to\ represent\ hyperbola\ is\ h^2\ -\ ab\ \gt\ 0\ \hspace{8cm}\]
\[Comparing\ with\ a\ x^2\ +\ 2h\ x\ y\ +\ by^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ = 0\]
\[We\ get\ a\ =\ 2,\ b\ =\ 8\ \hspace{5cm}\ 2h\ =\ – 16,\ \implies\ h\ =\ -\ 8\ \hspace{5cm}\]
\[\hspace{2cm}\ h^2\ -\ ab\ =\ (-8)^2\ -\ 2(4)\ =\ 64\ -\ 8\ =\ 56\ \gt\ 0\ \hspace{8cm}\]
\[\therefore\ (1)\ represents\ a\ hyperbola\ \hspace{10cm}\]
\[\color {purple} {4\ .}\ \color {red} {Find\ the\ equation\ of\ the\ parabola}\ with\ focus\ at\ (1,\ -1)\ \hspace{10cm}\]\[and\ directrix\ x\ -\ y\ =\ 0.\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ For\ parabola\ e\ =\ 1\ \hspace{15cm}\]
\[\hspace{2cm}\ Given\ Focus\ is\ S(1,\ -\ 1)\ and\ directrix\ is\ x\ -\ y\ =\ 0.\ \hspace{8cm}\]
\[Always\ \frac{SP}{PM}\ =\ e\ =\ 1\ \hspace{10cm}\]
\[\frac{\sqrt{(x\ -\ x_1)^2\ +\ (y\ -\ y_1)^2}}{\pm\ \frac{a\ x\ +\ b\ y\ +\ c}{\sqrt{a^2\ +\ b^2}}}\ =\ 1\ \hspace{10cm}\]
\[\frac{\sqrt{(x\ -\ 1)^2\ +\ (y\ +\ 1)^2}}{\pm\ \frac{x\ -\ y}{\sqrt{(1)^2\ +\ (-1)^2}}}\ =\ 1\ \hspace{10cm}\]
\[\sqrt{(x\ -\ 1)^2\ +\ (y\ +\ 1)^2}\ =\ \pm\ \frac{x\ -\ y}{\sqrt{2}}\ \hspace{10cm}\]
\[(x\ -\ 1)^2\ +\ (y\ +\ 1)^2\ =\ \frac{(x\ -\ y)^2}{2}\ \hspace{10cm}\]
\[\hspace{2cm}\ 2(x^2\ -\ 2\ x\ +\ 1\ +\ y^2\ +\ 2\ y\ +\ 1)\ =\ x^2\ +\ y^2\ -\ 2\ x\ y\ \hspace{8cm}\]
\[\hspace{2cm}\ 2\ x^2\ -\ 4\ x\ +\ 2\ +\ 2\ y^2\ +\ 4\ y\ +\ 2\ -\ x^2\ -\ y^2\ +\ 2\ x\ y\ =\ 0\ \hspace{10cm}\]
\[\hspace{2cm}\ 2\ x^2\ -\ 4\ x\ +\ 2\ +\ 2\ y^2\ +\ 4\ y\ +\ 2\ -\ x^2\ -\ y^2\ +\ 2\ x\ y\ =\ 0\ \hspace{8cm}\]
\[\hspace{2cm}\ \boxed {x^2\ +\ 2\ x\ y\ -\ 4\ x\ +\ y^2\ +\ 4\ y\ +\ 4\ =\ 0}\ \hspace{8cm}\]
\[\color {purple} {5\ .}\ \color {red} {Find\ the\ equation\ of\ the\ parabola}\ with\ focus\ at\ (2,\ 1)\ \hspace{10cm}\]\[and\ directrix\ 2x\ +\ y\ +\ 1\ =\ 0.\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ For\ parabola\ e\ =\ 1\ \hspace{15cm}\]
\[\hspace{2cm}\ Given\ Focus\ is\ S(2,\ 1)\ and\ directrix\ is\ 2\ x\ +\ y\ +\ 1\ =\ 0.\ \hspace{8cm}\]
\[Always\ \frac{SP}{PM}\ =\ e\ =\ 1\ \hspace{10cm}\]
\[\frac{\sqrt{(x\ -\ x_1)^2\ +\ (y\ -\ y_1)^2}}{\pm\ \frac{a\ x\ +\ b\ y\ +\ c}{\sqrt{a^2\ +\ b^2}}}\ =\ 1\ \hspace{10cm}\]
\[\frac{\sqrt{(x\ -\ 2)^2\ +\ (y\ -\ 1)^2}}{\pm\ \frac{2x\ +\ y\ +\ 1}{\sqrt{(2)^2\ +\ (1)^2}}}\ =\ 1\ \hspace{10cm}\]
\[\sqrt{(x\ -\ 2)^2\ +\ (y\ -\ 1)^2}\ =\ \pm\ \frac{2x\ +\ y\ +\ 1}{\sqrt{5}}\ \hspace{10cm}\]
\[(x\ -\ 2)^2\ +\ (y\ -\ 1)^2\ =\ \frac{(2x\ +\ y\ +\ 1)^2}{5}\ \hspace{10cm}\]
\[\hspace{2cm}\ 5(x^2\ -\ 4\ x\ +\ 4\ +\ y^2\ -\ 2\ y\ +\ 1)\ =\ 4x^2\ +\ y^2\ +\ 1\ +\ 4\ x\ y\ + 4\ x\ +\ 2y\ \hspace{8cm}\]
\[\hspace{2cm}\ 5\ x^2\ -\ 20\ x\ +\ 20\ +\ 5\ y^2\ -\ 10\ y\ +\ 5\ -\ 4\ x^2\ -\ y^2\ -\ 1\ -\ 4\ x\ y\ -\ 4x\ -\ 2y\ =\ 0\ \hspace{10cm}\]
\[\hspace{2cm}\ x^2\ +\ 4\ y^2\ -\ 4\ x\ y\ -\ 24\ x\ -\ 12\ y\ +\ 24\ =\ 0\ \hspace{8cm}\]
\[\hspace{2cm}\ \boxed {x^2\ +\ 4\ y^2\ -\ 4\ x\ y\ -\ 24\ x\ -\ 12\ y\ +\ 24\ =\ 0}\ \hspace{8cm}\]
\[\color {purple} {6.}\ \color {red} {What\ is\ the\ principal\ value\ of}\ Sin^{-1}\ (\frac{1}{2})\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[Sin^{-1}\ (\frac{1}{2})\ =\ 30^{0}\ \hspace{10cm}\]
\[\color {purple} {7.}\ \color {red} {What\ is\ the\ principal\ value\ of}\ Cos^{-1}\ (\frac{1}{2})\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[Cos^{-1}\ (\frac{1}{2})\ =\ 60^{0}\ \hspace{10cm}\]
\[\color {purple}{8.}\ \color {red}{Find\ the\ value\ of}\ i^2\ +\ i^3\ +\ i^4\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ i^2\ +\ i^3\ +\ i^4\ \hspace{19cm}\]
\[ =\ -1\ +\ i^2\ i\ +\ (i^2)^2\ \hspace{15cm}\]
\[ =\ -1\ +\ (-1)\ i\ +\ (-1)^2\ \hspace{15cm}\]
\[ =\ -1\ -\ i\ +\ 1\ \hspace{15cm}\]
\[ =\ -i\ \hspace{15cm}\]
\[\color {purple}{9.}\ If\ Z_1 = 1 + i,\ Z_2 = 3 + 2i,\ \color {red} {find\ Z_1 + Z_2}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ Z_1 = 1 + i,\ Z_2 = 3 + 2i\ \hspace{18cm}\]
\[Z_1 + Z_2 = 1 + i + 3 + 2i\ \hspace{10cm}\]
\[= 1 + 3 + i (1 +2)\ \hspace{10cm}\]
\[= 4 + 3i\ \hspace{10cm}\]
\[\color {purple} {10.}\ If\ Z_1 = 1\ +\ 4i,\ Z_2\ =\ -\ 3\ +\ 6i,\ \color {red} {find\ 3Z_1\ +\ 2\ Z_2}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ Z_1\ =\ 1\ +\ 4i,\ Z_2\ =\ -\ 3\ +\ 6i,\ \hspace{18cm}\]
\[3Z_1\ +\ 2\ Z_2\ =\ 3(1\ +\ 4i)\ +\ 2(-3\ +\ 6i)\ \hspace{10cm}\]
\[ = 3\ +\ 12i\ -\ 6\ +\ 12i\ \hspace{10cm}\]
\[= 3\ -\ 6\ + i (12\ +\ 12)\ \hspace{10cm}\]
\[=\ -3\ +\ 24i\ \hspace{10cm}\]
\[\color {purple} {11.}\ If\ Z_1 = 1 + i,\ Z_2 = 3 + 2i,\ \color {red} {find\ Z_1 – Z_2}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ Z_1 = 1 + i,\ Z_2 = 3 + 2i\ \hspace{18cm}\]
\[Z_1 – Z_2 = 1 + i – (3 + 2i)\ \hspace{10cm}\]
\[Z_1 – Z_2 = 1 + i – 3 – 2i\ \hspace{10cm}\]
\[Z_1 – Z_2 = 1 – 3 + i – 2i\ \hspace{10cm}\]
\[Z_1 – Z_2 = – 2 – i\ \hspace{10cm}\]
\[\color {purple} {12.}\ If\ Z_1 = 2\ +\ 2i,\ Z_2 =\ 3\ +\ 2i,\ \color {red} {find\ the\ value\ of\ z_1z_2}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ Z_1 = 2 + 2i,\ Z_2\ =\ 3\ +\ 2i\ \hspace{18cm}\]
\[Z_1 Z_2 = (2 + 2i) (3\ +\ 2i)\ \hspace{10cm}\]
\[Z_1 Z_2 = 6\ +\ 4i\ +\ 6i\ +\ 4 (-1)\ \hspace{5cm}\ \because i^2\ =\ -1\]
\[ = 6\ +\ 10i\ – 4\ \hspace{10cm}\]
\[ =\ 2\ +\ 10i\ \hspace{10cm}\]
\[\color {purple} {13.}\ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{2+ 3i}{4+ 5i}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ z = \frac{2+ 3i}{4+ 5i}\ ×\ \frac{4- 5i}{4- 5i}\ \hspace{18cm}\]
\[ = \frac{8 – 10i + 12i + 15}{(4)^2 + (5)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]\]
\[ = \frac{8 – 10i + 12i + 15}{16+ 25}\ \hspace{15cm}\]
\[ = \frac{23 + 2i}{41}\ \hspace{15cm}\]
\[ = \frac{23}{41}\ +\ i\ \frac{2}{41}\ \hspace{15cm}\]
\[Re(Z)\ =\ \frac{23}{41};\ Im(Z)\ =\ \frac{23}{41}\ \hspace{15cm}\]
\[\color {purple} {14.}\ \color {red} {Find\ the\ modulus\ and\ amplitude\ of}\ 1 + \sqrt{3}i\ \hspace{18cm}\]
\[\color {blue}{Solution:} \hspace{20cm}\]
\[Let\ z = 1 + \sqrt{3}i\ = a\ + ib\ \hspace{15cm}\]
\[a = 1,\ b\ = \sqrt{3}\ \hspace{15cm}\]
\[\color {brown} {T0\ find\ modulus}:\ \hspace{18cm}\]
\[|z| = \sqrt{a^2 + b^2}\ \hspace{12cm}\]
\[ = \sqrt{(1)^2 + (\sqrt{3})^2}\ \hspace{12cm}\]
\[ = \sqrt{1 + 3}\ =\ \sqrt{4}\ \hspace{12cm}\]
\[|z| = 2\ \hspace{15cm}\]
\[\color {brown} {To\ find\ amplitude}:\ \hspace{18cm}\]
\[θ = tan^{-1} (\frac{b}{a})\ =\ tan^{-1} \frac{\sqrt{3}}{1}\ =\ tan^{-1} \sqrt{3}\]
\[θ = 60^0\ \hspace{15cm}\]
\[\color {purple} {15.}\ \color {red} {Find\ the\ modulus\ and\ amplitude\ of}\ \frac{5 – i}{2- 3i}\ \hspace{18cm}\]
\[\color {blue}{Solution:} \hspace{22cm}\]
\[Let\ z = \frac{5 – i}{2- 3i}\ \hspace{18cm}\]
\[= \frac{5 – i}{2- 3i}\ ×\ \frac{2 + 3i}{2 + 3i}\ \hspace{15cm}\]
\[ = \frac{10 + 15i -2i – 3i^2}{(3)^2 + (2)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]\]
\[ = \frac{10 + 13i + 3 }{9+ 4}\ \hspace{4cm}\ \because [i^2\ =\ -1]\]
\[ = \frac{13 + 13i}{13}\ \hspace{10cm}\]
\[ Z = \frac{13}{13}\ +\ \frac{13i}{13}\ \hspace{15cm}\]
\[ z = 1 + \ i\ = a\ + ib\ \hspace{14cm}\]
\[a = 1,\ b\ = \ 1\ \hspace{15cm}\]
\[\color {brown} {T0\ find\ modulus}:\ \hspace{18cm}\]
\[|z| = \sqrt{a^2 + b^2}\ \hspace{12cm}\]
\[ = \sqrt{(1)^2 + {1}^2}\ \hspace{12cm}\]
\[ = \sqrt{1 + 1}\ =\ \sqrt{2}\ \hspace{12cm}\]
\[|z| = \sqrt{2}\ \hspace{15cm}\]
\[\color {brown} {To\ find\ amplitude}:\ \hspace{18cm}\]
\[θ = tan^{-1} (\frac{b}{a})\ =\ tan^{1} \frac{1}{1}\ =\ tan^{-1} {1}\]
\[θ = 45^0\ \hspace{15cm}\]
\[\color {purple} {16.}\ \color {red} {Simplify}\ (cos\ 15^0 + i sin\ 15^0 )^6\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ (cos\ 15^0 + i sin\ 15^0 )^6\ =\ cos\ 90^0 + i sin\ 90^0 \hspace{18cm}\]
\[= 0 + i (1)\ \hspace{10cm}\]
\[= i\ \hspace{10cm}\]
\[\boxed{(cos\ 15^0 + i sin\ 15^0 )^6\ =\ i}\]
\[\color {purple} {17.}\ \color {red} {Simplify:}\ (cos\ θ + i sin\ θ )^2\ (cos\ 3θ + i sin\ 3θ )\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ (cos\ θ + i sin\ θ )^2\ (cos\ 3θ + i sin\ 3θ )\ =\ (cos\ θ + i sin\ θ )^2\ (cos\ θ + i sin\ θ )^3\ \hspace{18cm}\]
\[= (cos\ θ + i sin\ θ )^{2 +3 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^5\ \hspace{10cm}\]
\[= cos\ 5θ + i sin\ 5θ\ \hspace{10cm}\]
\[\boxed{(cos\ θ + i sin\ θ )^2\ (cos\ 3θ + i sin\ 3θ )\ =\ cos\ 5θ + i sin\ 5θ}\]
\[\color {purple} {18.}\ If\ a = cos\ x + i sin\ x,\ b = cos\ y + i sin\ y,\ \color {red} {find\ ab\ and\ \frac{1}{ab}}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ a = cos\ x + i sin\ x,\ b = cos\ y + i sin\ y\ \hspace{16cm}\]
\[ab = (cos\ x + i sin\ x)\ (cos\ y + i sin\ y)\ \hspace{12cm}\]
\[\boxed{ab\ = cos\ (x + y) + i sin\ (x + y)}\ \hspace{10cm}\]
\[\frac{1}{ab}\ = \frac{1}{cos\ (x + y) + i sin\ (x + y)}\ \hspace{10cm}\]
\[\boxed{\frac{1}{ab}\ = cos\ (x + y) – i sin\ (x + y)}\ \hspace{10cm}\]
\[\color {purple}{19.}\ \color {red} {Simplify\ :}\ \frac{cos\ 5 θ + i sin\ 5 θ} {cos\ 3 θ + i sin\ 3 θ}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ \frac{cos\ 5 θ + i sin\ 5 θ} {cos\ 3 θ + i sin\ 3 θ}= \frac{(cos\ θ + i sin\ θ)^5} {(cos\ θ + i sin\ θ)^3} \hspace{18cm}\]
\[= (cos\ θ + i sin\ θ )^{5 -3 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^2\ \hspace{10cm}\]
\[= cos\ 2θ + i sin\ 2θ\ \hspace{10cm}\]
\[\boxed{\frac{cos\ 5 θ + i sin\ 5 θ} {cos\ 3 θ + i sin\ 3 θ}\ =\ cos\ 2θ + i sin\ 2θ}\]
\[\color {purple} {20.}\ \color {red} {Simplify\ using\ DeMoivre’s\ theorem:}\ \frac{(cos\ 3θ – i sin\ 3θ)^5\ (cos\ 4θ + i sin\ 4θ)^4} {(cos\ 2θ + i sin\ 2θ)^7\ (cos\ 3θ – i sin\ 3θ)^6}\ \hspace{10cm}\]
\[\color {blue}{Solution:}\ \frac{(cos\ 3θ – i sin\ 3θ)^5\ (cos\ 4θ + i sin\ 4θ)^4} {(cos\ 2θ + i sin\ 2θ)^7\ (cos\ 3θ – i sin\ 3θ)^6}\ \hspace{18cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^{5 \times -3}\ (cos\ θ + i sin\ θ)^{4 \times 4}} {(cos\ θ + i sin\ θ)^{2 \times 7}\ (cos\ θ + i sin\ θ)^{6 \times -3 }}\ \hspace{8cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^{-15}\ (cos\ θ + i sin\ θ)^{16}} {(cos\ θ + i sin\ θ)^{14}\ (cos\ θ + i sin\ θ)^{-18}}\ \hspace{8cm}\]
\[= (cos\ θ + i sin\ θ )^{-15 + 16 + 14 – 18 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^5\ \hspace{10cm}\]
\[= cos\ 5θ + i sin\ 5θ\ \hspace{10cm}\]
\[\boxed{\frac{(cos\ 3θ – i sin\ 3θ)^5\ (cos\ 4θ + i sin\ 4θ)^4} {(cos\ 2θ + i sin\ 2θ)^7\ (cos\ 3θ – i sin\ 3θ)^6}\ =\ cos\ 5θ + i sin\ 5θ}\]
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