Conic Sections

$E_1:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

  $e_1^2=1-\frac{b^2}{a^2}$ -(i)

Let $E_2:\frac{x^2}{a^2}+\frac{y^2}{B^2}=1$  

B > a

 $e_2^2=1-\frac{a^2}{B^2}andgivenB{e}_2=b$

$e^2=\frac{3-\sqrt[]{5}}{2}={\left(\frac{\sqrt[]{5}-1}{2}\right)}^2$

$\Rightarrow e=\frac{\sqrt[]{5}-1}{2}$

L : 2x + y = k.

$y=-2x+k.istangent\frac{x^2}{3}-\frac{y^2}{3}=1$                

$\Rightarrow k^2=3{\left(-2\right)}^2-3=9\Rightarrow k=3ask>0$              

y = -2x + 3 is also tangent to y² = $4\left(\frac{\alpha }{4}\right)x$  

$\Rightarrow 3=\frac{\alpha /4}{-2}$ -> a = -24

Centre of the smallest circle is A

Centre of the largest circle is B

$r_1=\left|CP+CB\right|=3\sqrt[]{2}+3$ and

$r_2=\left|CP-CB\right|=3\sqrt[]{2}-3$    

$\frac{r_1}{r_2}=\frac{3\sqrt[]{2}+3}{3\sqrt[]{2}-3}=3+2\sqrt[]{2}$

Let mid point of PQ is R (h, k)

$\therefore h=\frac{\alpha +\frac{4{\alpha }^2+\alpha +1}{2}}{2}and$

$k=\frac{4{\alpha }^2+1+\frac{4{\alpha }^2+\alpha +1}{2}}{2}$   

Eliminate a from above these two, we get

2 (3x – y)² + (x – 3y) + 2 = 0

  $\sim \left(\sim p\vee q\right)=p\wedge \sim q$

$AP=\sqrt[]{9+1-6+4+1}$

AP = 3 = AQ

$r=\sqrt[]{1+4-1}=2$        

$tan\theta =\frac{3}{2}$  

$\frac{Areaof\Delta APQ}{Areaof\Delta BPQ}=\frac{AR}{RB}=\frac{3sin\theta }{2cos\theta }=\frac{9}{4}$

The tangent to the parabola

$y^2=8xat\left(2,-4\right)is-4y=4\left(x+2\right)$  

x + y + 2 = 0

$OA=\sqrt[]{a}$         

$\Rightarrow \left|\frac{0+0+2}{\sqrt[]{2}}\right|=\sqrt[]{a}$  

$\sqrt[]{2}=\sqrt[]{a}$      

a = 2

Let point on ellipse (2sinθ, 3cosθ) and the mid point of line segment joining (-3, -5) and

(2 sin θ, 3cosθ) will be (h, k)

$\frac{2sin\theta -3}{2}=h\frac{3cos\theta -5}{2}=k$

sin² θ + cos² θ = 1

${\left(\frac{2h+3}{2}\right)}^2+{\left(\frac{2k+5}{3}\right)}^2$ = 1

$36x^2+16y^2+108x+80y+145=0$

$Area=\frac{1}{2}\left(\sqrt[]{5}-1\right)\frac{9-\sqrt[]{5}}{4}-{\displaystyle \underset{1}{\overset{\sqrt[]{5}}{\int }}\frac{1}{2}\sqrt[]{5-x^2}dx}$

$=\frac{1}{8}\left(-14+10\sqrt[]{5}\right)-\frac{1}{2}{\displaystyle \underset{co{s}^{-1}\frac{1}{\sqrt[]{5}}}{\overset{0}{\int }}\left(-si{n}^2\theta \right)d\theta }$

$A=\frac{-14}{8}+\frac{5}{4}\sqrt[]{5}-\left(\frac{5}{4}co{s}^{-1}\frac{1}{\sqrt[]{5}}-\frac{1}{2}\right)$

$=\frac{5}{4}\sqrt[]{5}-\frac{5}{4}-\frac{5}{4}co{s}^{-1}\frac{1}{\sqrt[]{5}}$

$\alpha =\frac{5}{4},\beta =\frac{-5}{4},\gamma =\frac{-5}{4}$

$\left|\alpha +\beta +\gamma \right|=\frac{5}{4}$

0 = -16 m + c . (i)

$\frac{\left|m\left(-10\right)+C\right|}{\sqrt[]{m^2+1}}=2.............\left(ii\right)$

(i) & (ii) Þ 36 m² = 4 (m² + 1)

$m=\frac{1}{2\sqrt[]{2}},C=\frac{8}{\sqrt[]{2}}$

$4\sqrt[]{2}\left(m+c\right)=34$