Straight Lines

D = |0, 2, 1; 1, -1, 1; x', y', 1| = 0

x² - 4xy – 5y² = 0
Equation of pair of straight line bisectors is (x²-y²)/ (a-b) = xy/h
(x²-y²)/ (1- (-5) = xy/ (-2)
(x²-y²)/6 = xy/ (-2)
x²-y² = -3xy
x² + 3xy - y² = 0

$h=\frac{cos\theta +3}{2}$

=> $k=\frac{sin\theta +2}{2}$

=> $cos\theta =2h-3\&sin\theta =2k-2$

${\left(h-3/2\right)}^2+{\left(k-1\right)}^2=\frac{1}{4}$

$\therefore$ circle of radius r = $\frac{1}{2}$

According to question,

$\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$           

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$  

Obviously B (2, 2) satisfying condition (i)

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 P (h, k)

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

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

  $8h^2+8k^2+100h+200=0$

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

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

$4r^2=\frac{225}{4}=56.25$         

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

$\therefore a=1,b=1,c=0$           

 $x=y^4,xy=k$

$\frac{dy}{dx}=\frac{1}{4y^3},\frac{dy}{dx}=\frac{-k}{x^2}$

P (x₁, y₁)

where $x_1=y_1^4\&x_1{y}_1=k$

$\Rightarrow y_1=k^{1/5},x_1{y}_1=k$

$m_1{m}_2=-1$

$\Rightarrow \frac{-1}{4.k^{6/5}}=-1\Rightarrow k^{6/5}=\frac{1}{4}\Rightarrow k^6=\frac{1}{4^5}=\frac{1}{2024}$

${\left(4k\right)}^6=2^{12}.\frac{1}{2^{10}}=2^2=4$

  $\frac{a+2+a}{3}=\frac{10}{3}$

2a + 2 = 0

2a = 8 -> a = 4       .(i)

and            $\frac{c+b+b}{3}=\frac{7}{3}$

2b + c = 7 .(ii)

Since         a, b, c are in A.P.

2b = a + c

From (i)   $\therefore$ 2b = 4 + c .(iii)

Solving (ii) and (iii)

4 + c + c = 7

2c = 3

$c=\frac{3}{2}$    

$\therefore 2b=4+\frac{3}{2}=\frac{11}{2}$     

$\therefore b=\frac{11}{4}$  

As per question

$\alpha +\beta =\frac{-b}{a}and\alpha \beta =\frac{1}{a}$       

$=\frac{121-192}{256}=\frac{-71}{256}$

K $=\frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}$  

K =    $\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$

$\therefore \frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}=\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$   

$\frac{x^2}{16}-\frac{y^2}{48}=1$

48 = 16 (e² – 1)

$e=\sqrt[]{4}=2$