Conic Sections

Equation of tangent is (x/a) (1/2) + (y/b) (√3/2) = 1
Equation of auxiliary circle is x² + y² = a²
Homogenising (ii) with (i) and making coefficient of x² + coefficient of y² = 0
⇒ (3a²/4b²) - (7/4) = 0 ⇒ e = 2/√7

Solving, x² – 9 = kx² ⇒ x² (k − 1) + 9 = 0 ⇒ x? + x? = 0 and x? = 9 / (k-1)
|x? - x? | = 10 = √ (x? + x? )² - 4x? x? ) ⇒ k = 16/25

Any point on x + y = 1 can be taken as (t, 1 – t) The equation of chord with this as mid-point is y (1 – t) -2a (x + t) = (1 – t)² – 4at

It passes through (a, 2a)

So, t² – 2t + 2a² – 2a + 1 = 0

This should have two distinct real roots.

So D > 0

$\begin{array}{l}\Rightarrow a^2-a<0\\ \Rightarrow 0<a<1\end{array}$          

So, length of latus rectum < 4 and 0 < a < 1

$\Rightarrow LR=1,2,3$            

PQ is focal chord.
Quadrilateral PTQR is square.

Area = (PQ * TR) / 2 = (4 * 4) / 2 = 8

Family of circle through (2, 2) and (9, 9), (x-2) (x-9) + (y-2) (y-9) + λ (y-x) = 0
Touches y=0 (x-axis)
(x-2) (x-9) + 18 - λx = 0
x² - 11x - λx + 36 = 0
x² - (11+λ)x + 36 = 0 has repeated roots
D = 0 ⇒ λ = 1, λ = -23
x = 6 or x = -6
Difference = 12

  $x^2+y^2-2x-6y+6=0$ centre (1, 3)

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

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

y² = 4x
(x - 3)² + y² = 9
y = mx + 1/m
(3, 0), r = 3

|3m + 1/m| / √ (1+m²) = 3

9m² + 1/m² + 6 = 9 (1+m²)
9m² + 1/m² + 6 = 9 + 9m²
1/m² = 3 ⇒ m = ±1/√3
m = 1/√3 in first quadrant
y = x/√3 + √3 ⇒ √3y = x + 3

x = 2t           $A\left(2t,\frac{t^2}{3}\right)$  

$y=\frac{t^2}{3}$          S (0, 3)

$3y={\left(\frac{x}{2}\right)}^2$           $B\left(0,\lambda \right)$  

$3k=\frac{t^2}{3}+3+\lambda =\frac{2t^2}{3}+3-\frac{12t^2}{9-t^2}$  

$\underset{t\to 1}{lim}3k=\frac{2}{3}+3-\frac{12}{8}=3-\frac{5}{6}=\frac{13}{6}$  

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

$x-ty+\frac{1}{4}{t}^2=0$

$\frac{1}{t}=m,m_2\Rightarrow \left(8m_1{m}_2\right)=3\sqrt[]{2}-4$

x² + y² = 36         ……(i)

y² = 9x      .(ii)

Solving (i) & (ii)

x² + 9x – 36 = 0

(x + 12) (x – 3) = 0

x = 3

Let $A={\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt[]{36-x^2}-3\sqrt[]{x}\right)dx}$  

$={\left[\frac{x\sqrt[]{36-x^2}}{2}+18si{n}^{-1}\frac{x}{6}\right]}_0^3-3.{\frac{x^{3/2}}{3/2}]}_0^3$   

  $=3\pi -\frac{3\sqrt[]{3}}{2}$           

$\therefore$ Required area = $=\frac{1}{2}\pi {\left(6\right)}^2+2\left(3\pi -\frac{3\sqrt[]{3}}{2}\right)=\left(24\pi -3\sqrt[]{3}\right)sq.unit$