Ncert Solutions Maths class 11th

Equation of normal is 4x – 3y + 1 = 0

Equation of tangent is 3x + 4y – 43 = 0

Area of triangle = $\frac{1}{2}\left(\frac{43}{3}+\frac{1}{4}\right)*7$  

$=\frac{1}{2}*\left(\frac{172+3}{12}\right)*7=\frac{1225}{24}$          

  $\therefore 24A=1225$        

Let a, ar, ar², ar³ are in G.P.

a + ar + ar² + ar³ = $\frac{65}{12}$  

$\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}+\frac{1}{a{r}^3}=\frac{65}{18}$        

${\left(ar\right)}^{2}r=\frac{3}{2}\Rightarrow a=\frac{2}{3},r=\frac{3}{2}$          

  $\therefore$   third term = ar² = $\frac{2}{3}*\frac{9}{4}=\frac{3}{2}$  

  $\therefore \alpha =\frac{3}{2}\Rightarrow 2\alpha =3$          

Let P (h, k)

$\sqrt[]{{\left(h-5\right)}^2+k^2}=3\sqrt[]{{\left(h+5\right)}^2+k^2}$           

$h^2-10h+25+k^2=9h^2+90h+225+9k^2$           

  $8h^2+8k^2+100h+200=0$

$x^2+y^2+\frac{25}{2}x+25=0$

$r^2=\frac{625}{16}-25=\frac{625-400}{16}=\frac{225}{16}$

$4r^2=\frac{225}{4}=56.25$         

${\left(x+1\right)}^2+\left|x-5\right|=\frac{27}{4}$

Case l        x < 5

$x^2+2x+1-x+5=\frac{27}{4}$       

$\left(2x+3\right)\left(2x-1\right)=0\Rightarrow x=-\frac{3}{2},\frac{1}{2}$     

Case II   $\ge$   x 5

$x^2+2x+1+x-5=\frac{27}{4}$       

$x=\frac{-12\pm \sqrt[]{144+43*16}}{2*4}=\frac{-12\pm \sqrt[]{832}}{8}\left(rejected\right)$           

           because x > 5

$L_1=\frac{x-\lambda }{1}=\frac{y-\frac{1}{2}}{\frac{1}{2}}=\frac{z}{-\frac{1}{2}}$           

$\therefore SD=\frac{\left|-2\lambda +3\left(2\lambda +\frac{1}{2}\right)+\lambda \right|}{\sqrt[]{14}}=\frac{\left|5\lambda +\frac{3}{2}\right|}{\sqrt[]{14}}$           

$5\lambda +\frac{3}{2}=-\frac{7}{2}\Rightarrow 5\lambda =-5\Rightarrow \lambda =-1$           

$\therefore \left|\lambda \right|=1$

S = 432 km/hr

tan 60° =   $\frac{h}{y}$

$h=\sqrt[]{3}$ y

$tan30°=\frac{h}{x+y}$           

$?x=432*\frac{1000}{3600}*20$           

x = 2400 m

y = 1200 m

$\therefore h=1200\sqrt[]{3}m$          

y = ax² + bx + c

a + b + c = 2

$\therefore \frac{dy}{dx}=2ax+b\Rightarrow {\frac{dy}{dx}|}_{\left(0,0\right)}=b=1$    

It passes through (0, 0)

0 = 0 + 0 + c -> c = 0

a = 1

$\therefore a=1,b=1,c=0$           

$tan\left(\frac{1}{4}si{n}^{-1}\frac{\sqrt[]{63}}{8}\right)$

$Put\frac{1}{4}si{n}^{-1}\frac{\sqrt[]{63}}{8}=\theta$

$sin4\theta =\frac{\sqrt[]{63}}{8}\therefore cos4\theta =\frac{1}{8}$

$co{s}^2\theta =\frac{7}{8}\Rightarrow se{c}^2\theta =\frac{8}{7}$

$ta{n}^2\theta =\frac{8}{7}-1=\frac{1}{7}$

$tan\theta =\frac{1}{\sqrt[]{7}}$

y = x² + 4

x² = y – 4

y = 4x – 1

$PQ=\frac{\left|4*\frac{1}{2}t-\frac{1}{4}{t}^2\right|-4-1}{\sqrt[]{17}}$           

$p=\frac{\left|t^2-8t-2\right|}{4\sqrt[]{17}}$          

$\frac{dp}{dt}=\frac{1}{4\sqrt[]{17}}$           (2t – 8) for maximum / minimum, $\frac{dp}{dt}=0\Rightarrow t=4$  

$also\frac{d^{2}p}{d{t}^2}>0att=4$           

Hence the closest point becomes at t = 4 is (2, 8)