Conic Sections

  $\frac{x^2}{4}+\frac{y^2}{2}=1$              

lim PQ is y = x + 1

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

$\Rightarrow {\left(x_1-x_2\right)}^2={\left(-\frac{4}{3}\right)}^2-4\left(-\frac{2}{3}\right)=\frac{16}{9}+\frac{8}{3}=\frac{40}{9}$               

y = x + 1

$y_1-y_2=x_1-x_2$

$P{Q}^2=\frac{80}{9}\Rightarrow \frac{PQ}{2}=r=\frac{\sqrt[]{80}}{6}$

$r=\frac{4\sqrt[]{5}}{6}\Rightarrow {\left(3r\right)}^2={\left(2\sqrt[]{5}\right)}^2=20$               

y = x – x²

p (t, t - t²)

$T_{\left(p\right)}:y+t-t^2$

$=x+t-2x.t$

$\Rightarrow y=\left(1-2t\right)x+t^2$

$\Rightarrow 1-2t=k,t^2=4,t=\pm 2$

$\Rightarrow y=5x+4$                

$x=\frac{1}{2}$

$y=2{\displaystyle \underset{y=0}{\overset{y=1}{\int }}\left(\frac{y^2+3}{4}-\frac{y^2+1}{2}\right)dy}$

$=\frac{1}{2}{\left(\frac{y^3}{3}+3y\right)}_0^1-{\left(\frac{y^3}{3}+y\right)}_0^1$

$=\frac{1}{2}\left(\frac{10}{3}\right)-\frac{4}{3}=\frac{1}{3}$

Let inclination of required line is,

So the coordinates of point B can be assumed as

$\left(4-\frac{\sqrt[]{29}}{3}cos\theta ,3-\frac{\sqrt[]{29}}{3}sin\theta \right)$

which satisfies x – y – 2 = 0

$sin\theta -cos\theta =\frac{3}{\sqrt[]{29}}$

By squaring

$sin2\theta =\frac{20}{29}=\frac{2tan\theta }{1+ta{n}^2\theta }$

Point B : $\left(\frac{10}{3},\frac{4}{3}\right)$ $\frac{}{}\frac{}{}$

which also satisfies x + 2y = 6

$\frac{{\left(x+1\right)}^2}{\lambda +5}=\frac{{\left(y+1\right)}^2}{\frac{\lambda +5}{4}}=1$  

length of latus rectum = $\frac{2b^2}{a}=\frac{2\left(\frac{\lambda +5}{4}\right)}{\sqrt[]{5+\lambda }}=4$  

$\sqrt[]{\lambda +5}=8\Rightarrow \lambda =59$                

Major axis = $2\sqrt[]{\lambda +5}=16$  

$e_H=\sqrt[]{1+\frac{64}{49}}=\frac{\sqrt[]{113}}{7}$

$\Rightarrow e_H.e_E=\frac{1}{2}$

$\Rightarrow \frac{113}{49}.\frac{\left(64-a^2\right)}{64}=\frac{1}{4}\Rightarrow a^2-64=\frac{32^2}{113}$

$\mathcal{l}=\frac{2a^2}{b}=2\left(64+\frac{32^2}{113}\right).\frac{1}{8}$

$113\mathcal{l}=1552$

Circle passes through (6, 1)

12 g – 19 c = 43               …. (i)

Centre lies on x – 2xy = 8

->g + 6c = 8                     …. (ii)

From (i) & (ii), c = 1, 9 = 2

Length of x – intercept -  $2\sqrt[]{g^2-C}$

tangent at (2t², 4t) is ty = x + 2t²,

It passes through (5, 7)

$2t^2-7t+5=0\Rightarrow t=1,\frac{5}{2}$                

$P\left(2t^2,4t\right)$ will be (2, 4), $\left(\frac{25}{2},0\right)$  

Let the equation of circle be

$x\left(x-\frac{1}{2}\right)+y^2+\lambda y=0$               

$\Rightarrow x^2+y^2-\frac{1}{2}x+\lambda y=0$

Radius = $\sqrt[]{\frac{1}{16}+\frac{{\lambda }^2}{4}}=2$  

$\Rightarrow {\lambda }^2=\frac{63}{4}\Rightarrow {\left(x-\frac{1}{4}\right)}^2+{\left(y+\frac{\lambda }{2}\right)}^2=4$   

$?$ This circle and parabola

$y-\alpha ={\left(x-\frac{1}{4}\right)}^2$ touch each other, so

$\alpha =-\frac{\lambda }{2}+2\Rightarrow \alpha -2=-\frac{\lambda }{2}\Rightarrow {\left(\alpha -2\right)}^2=\frac{{\lambda }^2}{4}=\frac{63}{16}$  

${\left(4\alpha -8\right)}^2=63$  

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{{\left(-4\sqrt[]{\frac{2}{5}}\right)}^2}{a^2}+\frac{32}{b^2}=1$                

$\Rightarrow \frac{32}{5a^2}+\frac{9}{b^2}=1$ ……. (i)

From (i)

$\frac{6}{b^2}+\frac{9}{b^2}=1\Rightarrow b^2=15\&a^2=16$

$a^2+b^2=15+16=31$