Chapter 10 - Conic Sections

Since the circle touches both axes in the first quadrant and radius is $a$, its centre is $(a, a)$.
Equation: $(x - a)^2 + (y - a)^2 = a^2$
$x^2 - 2ax + a^2 + y^2 - 2ay + a^2 = a^2$
$x^2 + y^2 - 2ax - 2ay + a^2 = 0$.

Calculate $x^2 + y^2$:
$x^2 + y^2 = \left(\frac{2at}{1+t^2}\right)^2 + \left(\frac{a(1-t^2)}{1+t^2}\right)^2$
$= \frac{a^2}{(1+t^2)^2} [4t^2 + (1-t^2)^2]$
$= \frac{a^2}{(1+t^2)^2} [4t^2 + 1 - 2t^2 + t^4]$
$= \frac{a^2}{(1+t^2)^2} [t^4 + 2t^2 + 1]$
$= \frac{a^2(1+t^2)^2}{(1+t^2)^2} = a^2$.
Thus, the point lies on the circle $x^2 + y^2 = a^2$.

Let the equation be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Passes through $(0, 0) \implies c = 0$.
Passes through $(a, 0) \implies a^2 + 2ga = 0 \implies g = -a/2$.
Passes through $(0, b) \implies b^2 + 2fb = 0 \implies f = -b/2$.
Centre is $(-g, -f) = (a/2, b/2)$.

Centre $(1, 2)$. Since it touches the x-axis, radius $r = |y\text{-coordinate}| = 2$.
Equation: $(x - 1)^2 + (y - 2)^2 = 2^2$
$x^2 - 2x + 1 + y^2 - 4y + 4 = 4$
$x^2 + y^2 - 2x - 4y + 1 = 0$.

The lines are parallel. $3x - 4y + 4 = 0$ and $3x - 4y - 7/2 = 0$.
Distance between them is the diameter.
$D = \frac{|4 - (-3.5)|}{\sqrt{3^2 + (-4)^2}} = \frac{7.5}{5} = 1.5$.
Radius $r = D/2 = 0.75 = 3/4$.

Since it lies in the 3rd quadrant and touches both axes, centre is $(-r, -r)$ and radius is $r$.
Distance from centre to line $3x - 4y + 8 = 0$ is $r$.
$\frac{|3(-r) - 4(-r) + 8|}{\sqrt{3^2 + (-4)^2}} = r$
$\frac{|r + 8|}{5} = r \implies |r + 8| = 5r$.
$r + 8 = 5r \implies 4r = 8 \implies r = 2$.
($r + 8 = -5r$ gives negative radius, impossible).
Centre $(-2, -2)$, radius 2.
Equation: $(x + 2)^2 + (y + 2)^2 = 4 \implies x^2 + y^2 + 4x + 4y + 4 = 0$.

Centre of circle is $(2, 3)$.
Let other end be $(x, y)$. Midpoint of diameter is centre.
$\frac{x + 3}{2} = 2 \implies x = 1$.
$\frac{y + 4}{2} = 3 \implies y = 2$.
Coordinates are $(1, 2)$.

Intersection of lines: $y = 14 - 3x$.
$2x + 5(14 - 3x) = 18 \implies 2x + 70 - 15x = 18 \implies -13x = -52 \implies x = 4$.
$y = 14 - 12 = 2$. Point is $(4, 2)$.
Radius $r = \sqrt{(4-1)^2 + (2-(-2))^2} = \sqrt{9 + 16} = 5$.
Equation: $(x - 1)^2 + (y + 2)^2 = 25 \implies x^2 + y^2 - 2x + 4y - 20 = 0$.

Line is $\sqrt{3}x - y + k = 0$. Circle centre $(0, 0)$, radius 4.
Perpendicular distance from centre to line equals radius.
$\frac{|k|}{\sqrt{(\sqrt{3})^2 + (-1)^2}} = 4 \implies \frac{|k|}{2} = 4 \implies |k| = 8 \implies k = \pm 8$.

Given circle centre $(3, -6)$, radius $r_1 = \sqrt{9 + 36 - 15} = \sqrt{30}$.
Area $A_1 = 30\pi$. New area $A_2 = 60\pi$. New radius $r_2 = \sqrt{60}$.
Equation: $(x - 3)^2 + (y + 6)^2 = 60$
$x^2 - 6x + 9 + y^2 + 12y + 36 = 60$
$x^2 + y^2 - 6x + 12y - 15 = 0$.

Latus rectum $\frac{2b^2}{a}$. Minor axis $2b$.
$\frac{2b^2}{a} = \frac{1}{2}(2b) = b \implies 2b = a$.
$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{b^2}{4b^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.

Equation: $\frac{x^2}{25} + \frac{y^2}{9} = 1$. $a = 5, b = 3$.
$e = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Foci $(\pm ae, 0) = (\pm 5(4/5), 0) = (\pm 4, 0)$.

$2ae = 10 \implies ae = 5$. Given $e = 5/8$.
$a(5/8) = 5 \implies a = 8$.
$b^2 = a^2(1 - e^2) = 64(1 - 25/64) = 64 - 25 = 39$.
Latus rectum = $\frac{2b^2}{a} = \frac{2(39)}{8} = \frac{39}{4}$.

$e = 2/3$. $\frac{2b^2}{a} = 5 \implies b^2 = \frac{5a}{2}$.
$b^2 = a^2(1 - e^2) \implies \frac{5a}{2} = a^2(1 - 4/9) = a^2(5/9)$.
$\frac{1}{2} = \frac{a}{9} \implies a = 4.5 = 9/2$.
$b^2 = \frac{5(9/2)}{2} = \frac{45}{4}$.
Equation: $\frac{x^2}{81/4} + \frac{y^2}{45/4} = 1 \implies \frac{4x^2}{81} + \frac{4y^2}{45} = 1$.

$a^2 = 36, b^2 = 20$. $e = \sqrt{1 - 20/36} = \sqrt{16/36} = 4/6 = 2/3$.
Directrices $x = \pm a/e = \pm 6/(2/3) = \pm 9$.
Distance = $9 - (-9) = 18$.

$4a = 8 \implies a = 2$. Focus $(2, 0)$.
Focal distance of $(x, y)$ is $x + a$.
$x + 2 = 4 \implies x = 2$.
$y^2 = 8(2) = 16 \implies y = \pm 4$.

Let point be $P(at^2, 2at)$. Vertex $O(0, 0)$.
Slope of $OP = \tan \theta = \frac{2at}{at^2} = \frac{2}{t} \implies t = 2 \cot \theta$.
Length $OP = \sqrt{(at^2)^2 + (2at)^2} = at\sqrt{t^2 + 4}$
$= a(2\cot\theta)\sqrt{4\cot^2\theta + 4} = 2a\cot\theta \cdot 2\csc\theta = 4a \frac{\cos\theta}{\sin^2\theta} = 4a \cos\theta \csc^2\theta$.

Vertex $V(0, 4)$, Focus $F(0, 2)$. Axis is y-axis.
Distance $a = VF = 2$. Since focus is below vertex, parabola opens downwards.
Equation: $(x - 0)^2 = -4a(y - 4) \implies x^2 = -8(y - 4)$.

Condition for tangency for $y = mx + c$ to $y^2 = 4ax$ is $c = a/m$.
Here $a = 1, c = 1$.
$1 = 1/m \implies m = 1$.

$2ae = 16 \implies ae = 8$. Given $e = \sqrt{2}$.
$a\sqrt{2} = 8 \implies a = 4\sqrt{2} \implies a^2 = 32$.
$b^2 = a^2(e^2 - 1) = 32(2 - 1) = 32$.
Equation: $\frac{x^2}{32} - \frac{y^2}{32} = 1 \implies x^2 - y^2 = 32$.

Equation: $\frac{y^2}{4} - \frac{x^2}{9} = 1$. This is a vertical hyperbola.
$a^2 = 4, b^2 = 9$.
$c^2 = a^2 + b^2 = 4 + 9 = 13$.
$e = \frac{c}{a} = \frac{\sqrt{13}}{2}$.

Foci $(\pm ae, 0) = (\pm 2, 0) \implies ae = 2$.
$e = 3/2 \implies a(3/2) = 2 \implies a = 4/3 \implies a^2 = 16/9$.
$b^2 = a^2(e^2 - 1) = \frac{16}{9}(\frac{9}{4} - 1) = \frac{16}{9}(\frac{5}{4}) = \frac{20}{9}$.
Equation: $\frac{x^2}{16/9} - \frac{y^2}{20/9} = 1$.