Conic Sections

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 = $\frac{\sqrt[]{5}}{3}$

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

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

sin = $\frac{AP}{AN}$

 $AP\perp BP$

M₁ M₂ = 1

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

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

Equation of tangent of slope m to y = x² is y = mx $\frac{1}{4}{m}^2$ - …………. (i)

Equation of tangent of slope m to y = - (x - 2)² is y = m (x – 2) + $\frac{1}{4}{m}^2$  …………… (ii)

If both equation represent the same line therefore on comparing (i) and (ii) we get m = 0, 4

therefore equation of tangent is y = 4x – 4

 $\frac{x^2}{{\left(\sqrt[]{\frac{5}{2}}\right)}^2}+\frac{y^2}{{\left(\sqrt[]{\frac{5}{3}}\right)}^2}=1$

Equation of tangent having slope m is

$y=mx\pm \sqrt[]{\frac{5}{3}{m}^2+\frac{5}{3}},$ which passes through (1, 3) and we get m₁ + m₂ = -4 and m₁m₂ = $\frac{44}{9}$

$\therefore$ Acute angle between the tangents is α  = tan^-1 $\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=ta{n}^{-1}\left(\frac{24}{7\sqrt[]{5}}\right)$

 $y^2=\frac{-x}{2}$

$y=mx-\frac{1}{8m}$

This tangent pass through (2, 0)

$m=\pm \frac{1}{4}i.e.,onetangentisx-4y-2=0,17r=9$

Ellipse is $\frac{x^2}{4}+\frac{y^2}{4}=1e=\frac{1}{\sqrt[]{2}}S\equiv \left(0,-\sqrt[]{2}\right)$

Chord of contact is

$\frac{x}{\sqrt[]{2}}+\frac{\left(2\sqrt[]{2}-2\right)y}{4}=1$

$P\equiv \left(1,\sqrt[]{2}\right)Q\equiv \sqrt[]{2},0$

$\therefore {\left(SP\right)}^2+{\left(SQ\right)}^2=13$

$H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Foci : S (ae, 0), S' (ae, 0)

Focus of parabola is (ae, 0)

Now, semi latus rectum of parabola = |SS'| = 2ae

$Given,4ae=e\left(\frac{2b^2}{a}\right)$

B² = 2a²……… (i)

$Given,\left(2\sqrt[]{2},-2\sqrt[]{2}\right)$ lies on H

$\Rightarrow \frac{1}{a^2}-\frac{1}{b^2}=\frac{1}{8}........(ii)$

From (i) and (ii)

$a^2=4,b^2=8$

$?b^2=a^2\left(e^2-1\right)$

$\therefore e=\sqrt[]{3}$

$\Rightarrow$ equation of parabola is  y² $=8\sqrt[]{3}x$