Conic Sections

$x^2+y^2-2x-6y+6=0$  centre (1, 3)

$\begin{array}{l}r=\sqrt[]{1+9-6}=2\\ CM=\sqrt[]{1+4}=\sqrt[]{5}\end{array}$

$r=\sqrt[]{5+4}=3$

$M=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill c_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]$    

$M^{T}M=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill b_1\hfill & \hfill c_1\hfill \\ \hfill a_2\hfill & \hfill b_2\hfill & \hfill c_2\hfill \\ \hfill a_3\hfill & \hfill b_3\hfill & \hfill c_3\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]$

$Tr\left(M^{T}M\right)=a_1^2+b_1^2+c_1^2+a_2^2+b_2^2+c_2^2+a_3^2+b_3^2+c_3^2=7$           

all   $a_i,b_i,c_i\in \left\{0,1,2\right\}for$ i = 1, 2, 3

Case 1 7 one's and two zeroes which can occur in ways

Case 2 One 2 three 1's five zeroes =

total such matrices = 504 + 36 = 540

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)

${=}^{n+1}{C}_2+2{\displaystyle \sum _{r=1}^{n-1}\frac{\left(r+1\right)!}{\left(r-1\right)!2!}}{=}^{n+1}{C}_2+{\displaystyle \sum _{r=1}^{n-1}\left(r+1\right)r}$

  =   $\frac{\left(n+1\right)!}{2!\left(n+1\right)!}+\frac{\left(n-1\right)n\left(2\left(n-1\right)+1\right)}{6}+\frac{\left(n-1\right)n}{2}=\frac{n\left(n+1\right)}{2}+\frac{n\left(n-1\right)\left(2n-1\right)}{6}+\frac{n\left(n-1\right)}{2}$

=   $\frac{n\left(6n+2n^2-3n+1\right)}{6}=\frac{\left(2n^2+3n+1\right)}{6}=\frac{n\left(2n+1\right)\left(n+1\right)}{6}$

$x+\sqrt[]{3}y=2\sqrt[]{3},andm=-\frac{1}{\sqrt[]{3}},C=2$          not possible

 (2) , $\left(1\right)\frac{x^2}{\frac{9}{2}}-\frac{y^2}{\frac{1}{2}}=1\Rightarrow c=\sqrt[]{a^2{m}^2-b^2}=\sqrt[]{\frac{9}{2}*\frac{1}{3}-\frac{1}{2}}=1,$ not possible

(3) $c=a\sqrt[]{1+m^2}=\sqrt[]{7}*2=2\sqrt[]{7},$ not possible

(4) $\frac{x^2}{9}+\frac{y^2}{1}=1,C=\sqrt[]{9m^2+1}=\sqrt[]{9*\frac{1}{3}+1}=2$  

A tangent to y² = 4x is x – ty + t² = 0

$\frac{3+t^2}{\sqrt[]{1+t^2}}=3$

(3 + t²)² = 9 (1 + t²)

$9+t^4+6t^2=9+9t^2$

Point of contact $\left(3,2\sqrt[]{3}\right)=\left(a,b\right)$

$x-\sqrt[]{3}y+3=0$

$\frac{\sqrt[]{3}x+y-3\sqrt[]{3}=0]}{4x-6=0}$

$x=\frac{3}{2},y=\frac{\sqrt[]{3}}{2}+\sqrt[]{3}$

& $\left(\frac{3}{2},\frac{\sqrt[]{3}}{2}+2\right)=\left(c,d\right),2\left(a+c\right)=2\left(3+\frac{3}{2}\right)=9$

2x + y = 1

$m_1=\frac{-2}{1}=-2and{m}_1{m}_2=-1$          

$-2.m_2=-1$           

$m_2=\frac{1}{2}$           

y² = 6x

y² = 4 (3/2) x

$y=mx+\frac{a}{m}$         

$y=\frac{1}{2}x+\frac{\frac{3}{2}}{\frac{1}{2}}$        

$y=\frac{x}{2}+3$           

$\frac{x^2}{25}+\frac{y^2}{16}=1$

16 = 25 (1 – e²)

$\Rightarrow e=\frac{3}{5}$

OF₁ = 5 $\left(\frac{3}{5}\right)=3$

For Hyperbola : e' $=\frac{5}{3}$

a = 3

$Hyperbola,\frac{x^2}{9}-\frac{y^2}{16}=1$

Combined equation of pair of lines OP and OQ is

$x^2+2y^2=2{\left(x+y\right)}^2$

$\Rightarrow x^2+4xy=0\Rightarrow x\left(x+4y\right)=0\Rightarrow \{\begin{array}{l}x=0\left(lineOP\right)\\ y=-\frac{x}{4}\left(lineOQ\right)\end{array}$

$tan\left(90°+\theta \right)=-\frac{1}{4}$

$-cot\theta =-\frac{1}{4}\Rightarrow tan\theta =4$

$\theta =ta{n}^{-1}4=co{t}^{-1}\frac{1}{4}=\frac{\pi }{2}-ta{n}^{-1}\frac{1}{4}$

$\angle POQ=\pi -\theta =\frac{\pi }{2}+ta{n}^{-1}\frac{1}{4}$

Using the standard equations of a hyperbola:

$9\left({\mathrm{e}}^2+l\right)$ and directrix $focus\mathrm{a}\mathrm{e}=\sqrt[]{10}$

By multiplying both focus and directrix, we get
$\frac{\mathrm{a}}{\mathrm{e}}=\frac{9}{\sqrt[]{10}}$ and $\Rightarrow {\mathrm{a}}^2=9$
Now $\mathrm{e}=\frac{\sqrt[]{10}}{3}$
$\mathrm{}(ae{)}^2=a^2+b^2$