What are the Key topics discussed in the Conic Sections chapter?

Given 2a + 2b = $4\left(2\sqrt[]{2}+\sqrt[]{14}\right)$

$=4\sqrt[]{2}\left(2+\sqrt[]{7}\right)$

$b^2=a^2\left(\frac{11}{4}-1\right)$

4b2 = 7a2

b = $\frac{\sqrt[]{7}}{2}a$

Tangent to C₁ at (-1, 1) is T = 0                                                           

 x(-1) + 4(1) = 2

-x + y = 2

find OP by dropping  from (3, 2) to centre

OP = $\left|\frac{3-2+2}{\sqrt[]{2}}\right|=\frac{3}{\sqrt[]{2}}$

AP = $\sqrt[]{r^2-O{P}^2}$

$=\sqrt[]{5-\frac{9}{2}}=\frac{1}{\sqrt[]{2}}$

$tan\theta =\frac{OP}{AP}=\frac{3/\sqrt[]{2}}{1/\sqrt[]{2}}=3$

area of $\Delta ABN=\frac{1}{2}A{N}^2\cdot sin2\theta$

AN = $\frac{\sqrt[]{5}}{3}$

$=\frac{1}{2}\cdot \frac{5}{9}\cdot \left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)$

$=\frac{5}{2.9}\times \frac{2.3}{1+3^2}=\frac{1}{6}$

sin = $\frac{AP}{AN}$

 $AP\perp BP$

M₁ M₂ = 1

$\frac{2t}{t^2-3}\times \frac{2/6}{3-\frac{1}{t^2}}=-1$

t = 1

So, A (1, 2) and B (1, 2) they must be end pts of focal chord.

Length of latus rectum $=\frac{2b^2}{a}$

$4=\frac{2b^2}{a}$

b2 = 2a and ae = 1

Eccentricity of ellipse (Horizontal)

b² = a² (1 – e²)

2a = a² (1 – e²)

2 = $\frac{1}{e}\left(1-e^2\right)$

e² + 2e – 1 = 0

$e=\frac{-2\pm \sqrt[]{4+4}}{2}$

$e=-1+\sqrt[]{2}$

now $\frac{1}{e^2}$$=$$3$$+$$2$

Given hyperbola : $\frac{k{x}^2}{6}-\frac{y^2}{6}=1$ so eccentricity e = $\sqrt[]{1+k}$ and directrices $x=\pm \frac{a}{e}$

$\Rightarrow x=\pm \frac{\sqrt[]{6}}{\sqrt[]{k}\sqrt[]{k+1}}\Rightarrow \frac{\sqrt[]{6}}{\sqrt[]{k}\sqrt[]{k+1}}=1$

k = 2 therefore equation of hyperbola is $\frac{x^2}{3}-\frac{y^2}{6}=1$

hence it passes through the point $\left(\sqrt[]{5},-2\right)$

Abscissae of PQ are roots of x² – 4x – 6 = 0

Ordinates of PQ are roots of y² + 2y – 7 = 0 and PQ is diameter

$\therefore$ Equation of circle is $x^2+y^2-4x+2y-13=0$   ……………. (i)

But, given  $x^2+y^2+2ax+2by+c=0$ ……………. (ii)

By comparison a = -2, b = 1, c = -13 Þ a + b – c = -2 + 1 + 13 = 12