Ncert Solutions Maths class 11th

Let height of the wall = h than A(0, 0, 0) G (10, 10, h)

$\overrightarrow{AG}=10\widehat{i}+10\widehat{j}+h\widehat{k}$              

B(10, 0, 0) A (0, 10, h)

$\overrightarrow{BH}=-10\widehat{i}+10\widehat{j}+h\widehat{k}$              

  $cos\theta =\frac{1}{5}=\frac{\overrightarrow{BH}.\overrightarrow{AG}}{\left|\overrightarrow{BH}\right|\left|\overrightarrow{AG}\right|}$             

$=\frac{-100+100+h^2}{\left(200+h^2\right)}$              

$5h^2=200+h^2\Rightarrow h^2=50$              

$h=5\sqrt[]{2}$

${\left(\sqrt[]{3}+i\right)}^{100}=2^{100}{\left(\frac{\sqrt[]{3}}{2}+i\frac{1}{2}\right)}^{100}$

$=2^{100}{\left(cos\frac{\pi }{6}+isin\frac{\pi }{6}\right)}^{100}$

$=2^{99}\left(-1+i\sqrt[]{3}\right)=2^{99}\left(p+iq\right)$

$\therefore p=-1\&q=\sqrt[]{3}$              

$\therefore$ Required equation $x^2-\left(\sqrt[]{3}-1\right)x-\sqrt[]{3}=0$

Equation of required circle

$C:{\left(x-2\right)}^2+{\left(y-1\right)}^2+\lambda \left(2y-x\right)=0..........\left(i\right)$              

Intersect C₁x² + y² + 2y – 5 = 0 .(ii)

$\therefore$ Equation of radical axis

    $-4x-4y+10+\lambda \left(2y-x\right)=0..........\left(ii\right)$          

Centre of C₁ (0, -1) lies on .(ii)

$\Rightarrow 4+10-2\lambda =0\Rightarrow \lambda =7$        

Equation of circle C is

$x^2+y^2-11x+12y+5=0$       

Diameter = $\sqrt[]{245}=7\sqrt[]{5}$  

$\frac{a}{1-r}=15,\frac{a^2}{1-r^2}=150$

$\Rightarrow r=\frac{1}{5},a=12\Rightarrow \frac{a{r}^2}{1-r^2}=\frac{1}{2}$

Line BM : 2x + y = 3 Þ M (0, 3)

Line CD : 7x – 4y = 1 Þ C (3, 5)

Mirror image of A (-3, 1) in the line CD is $\left(\frac{13}{5},\frac{-11}{5}\right)$  and it will lie on BC.

Slope of AC is 2/3

Slope of BC is 18.

$tan\theta =\frac{18-\frac{2}{3}}{1+18\left(\frac{2}{3}\right)}=\frac{4}{3}$   

 where θ = $\angle ACB$  

Kindly consider the following figure

$P\left(\frac{15}{2}{t}^2,15t\right)$

C (-15,0)

Q (-30,0)

tanθ = $\frac{PN}{QN}$

=> $\frac{1}{t}=\frac{15t}{30+\frac{15}{2}{t}^2}$

=> t = 2

$Tangent=y=\frac{1}{2}\left(x+30\right)$

$T_{\left(P\right)}:\frac{x{x}_1}{8}+\frac{y{y}_1}{4}=1,$

where $\frac{x_{1}2}{8}+\frac{y_{1}2}{4}=1...........\left(i\right)$

given : slope = 2

=> x₁ = -4y         . (ii)

(i) & (ii) => P $\left(-\frac{8}{3},\frac{2}{3}\right)$

$e=\frac{1}{\sqrt[]{2}}$         

A = ar (PSS') = $\frac{4}{3}$

$\left(5-e^2\right)A=6$

$S-\frac{1}{x-1}=\frac{-2^{101}}{{\left(x^{2^{100}}\right)}^2-1}$

For x = 2, S = 1 $\frac{2^{101}}{4^{101}-1}$

Centre (0, 1)

Radius = $\sqrt[]{2}$