Chapter 10: Conic Sections

We are given $h = 0$, $k = 2$, and $r = 2$. Substituting into the standard circle equation $(x-h)^2 + (y-k)^2 = r^2$:

$$(x-0)^2 + (y-2)^2 = 2^2$$

Expanding and simplifying:

$$x^2 + y^2 – 4y + 4 = 4$$ $$\boxed{x^2 + y^2 – 4y = 0}$$

We are given $h = -2$, $k = 3$, and $r = 4$. Using the standard form $(x-h)^2 + (y-k)^2 = r^2$:

$$(x+2)^2 + (y-3)^2 = 4^2$$

Expanding:

$$x^2 + 4x + 4 + y^2 – 6y + 9 = 16$$ $$\boxed{x^2 + y^2 + 4x – 6y – 3 = 0}$$

The standard equation of a circle with centre $(h, k)$ and radius $r$ is:

$$(x-h)^2 + (y-k)^2 = r^2$$

Here $(h, k) = \left(\dfrac{1}{2}, \dfrac{1}{4}\right)$ and $r = \dfrac{1}{12}$. Substituting these values:

$$\left(x-\frac{1}{2}\right)^2 + \left(y-\frac{1}{4}\right)^2 = \left(\frac{1}{12}\right)^2$$

Expanding the left side:

$$x^2 + \frac{1}{4} – x + y^2 + \frac{1}{16} – \frac{1}{2}y = \frac{1}{144}$$ $$x^2 + y^2 – x – \frac{1}{2}y + \frac{11}{36} = 0 \quad \left[\because \frac{1}{4}+\frac{1}{16}-\frac{1}{144}=\frac{36+9-1}{144}=\frac{11}{36}\right]$$

Multiplying throughout by 36:

$$\boxed{36x^2 + 36y^2 – 36x – 18y + 11 = 0}$$

We are given $h = 1$, $k = 1$, and $r = \sqrt{2}$. Using the standard form:

$$(x-1)^2 + (y-1)^2 = (\sqrt{2})^2$$

Expanding:

$$x^2 – 2x + 1 + y^2 – 2y + 1 = 2$$ $$\boxed{x^2 + y^2 – 2x – 2y = 0}$$

We are given $h = -a$, $k = -b$, and $r = \sqrt{a^2 – b^2}$. Substituting into the standard form:

$$(x+a)^2 + (y+b)^2 = a^2 – b^2$$

Expanding the left side:

$$x^2 + 2ax + a^2 + y^2 + 2by + b^2 = a^2 – b^2$$ $$\boxed{x^2 + y^2 + 2ax + 2by + 2b^2 = 0}$$

We rewrite the given equation as $[x-(-5)]^2 + (y-3)^2 = 6^2$. Comparing with the standard form $(x-h)^2 + (y-k)^2 = r^2$:

$$h = -5,\quad k = 3,\quad r = 6$$

∴ The circle has centre $(-5,\, 3)$ and radius $6$.

We group the $x$ and $y$ terms and move the constant to the right:

$$(x^2 – 4x) + (y^2 – 8y) = 45$$

Now we complete the square for both variables by adding $\left(\frac{1}{2}\text{ coeff. of }x\right)^2$ and $\left(\frac{1}{2}\text{ coeff. of }y\right)^2$ to both sides:

$$(x^2 – 4x + 2^2) + (y^2 – 8y + 4^2) = 45 + 4 + 16$$ $$(x-2)^2 + (y-4)^2 = 65$$

Comparing with $(x-h)^2 + (y-k)^2 = r^2$:

$$h = 2,\quad k = 4,\quad r = \sqrt{65}$$

∴ The circle has centre $(2,\, 4)$ and radius $\sqrt{65}$.

Rearranging by grouping $x$ and $y$ terms:

$$(x^2 – 8x) + (y^2 + 10y) = 12$$

Completing the square on both sides (adding $4^2 = 16$ and $5^2 = 25$):

$$(x^2 – 8x + 4^2) + (y^2 + 10y + 5^2) = 12 + 16 + 25$$ $$(x-4)^2 + (y+5)^2 = 53$$ $$(x-4)^2 + [y-(-5)]^2 = (\sqrt{53})^2$$

Comparing with the standard form:

$$h = 4,\quad k = -5,\quad r = \sqrt{53}$$

∴ The circle has centre $(4,\, -5)$ and radius $\sqrt{53}$.

We divide every term by 2 so that the coefficients of $x^2$ and $y^2$ become 1:

$$x^2 + y^2 – \frac{1}{2}x = 0$$ $$\left(x^2 – \frac{1}{2}x\right) + y^2 = 0$$

Completing the square: add $\left(\frac{1}{4}\right)^2 = \frac{1}{16}$ to both sides:

$$\left(x^2 – \frac{1}{2}x + \left(\frac{1}{4}\right)^2\right) + y^2 = \frac{1}{16}$$ $$\left(x – \frac{1}{4}\right)^2 + y^2 = \left(\frac{1}{4}\right)^2$$

Comparing with the standard form:

$$h = \frac{1}{4},\quad k = 0,\quad r = \frac{1}{4}$$

∴ The circle has centre $\left(\dfrac{1}{4},\, 0\right)$ and radius $\dfrac{1}{4}$.

Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$  …(i)

Since $(4,1)$ and $(6,5)$ lie on the circle:

$$(4-h)^2 + (1-k)^2 = r^2 \quad\cdots(ii)$$ $$(6-h)^2 + (5-k)^2 = r^2 \quad\cdots(iii)$$

Setting (ii) = (iii) and expanding:

$$-8h – 2k + 17 = -12h – 10k + 61 \Rightarrow 4h + 8k = 44$$ $$\Rightarrow h + 2k = 11 \quad\cdots(iv)$$

Since the centre $(h,k)$ lies on $4x + y = 16$:

$$4h + k = 16 \quad\cdots(v)$$

Solving (iv) and (v): from $(v) – 4\times(iv)$: $k – 8k = 16 – 44 \Rightarrow k = 4$.

Substituting $k = 4$ in (iv): $h + 8 = 11 \Rightarrow h = 3$.

Finding $r^2$ from (ii): $r^2 = (4-3)^2 + (1-4)^2 = 1 + 9 = 10$.

The equation of the required circle:

$$x^2 + 9 – 6x + y^2 + 16 – 8y = 10$$ $$\boxed{x^2 + y^2 – 6x – 8y + 15 = 0}$$

Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$  …(i)

Since $(2,3)$ and $(-1,1)$ lie on the circle:

$$(2-h)^2 + (3-k)^2 = r^2 \quad\cdots(ii)$$ $$(-1-h)^2 + (1-k)^2 = r^2 \quad\cdots(iii)$$

Setting (ii) = (iii) and simplifying:

$$13 – 4h – 6k = 2 + 2h – 2k \Rightarrow 6h + 4k = 11 \quad\cdots(iv)$$

Since the centre $(h,k)$ lies on $x – 3y – 11 = 0$:

$$h – 3k = 11 \quad\cdots(v)$$

From $(iv) – 6\times(v)$: $4k + 18k = 11 – 66 \Rightarrow 22k = -55 \Rightarrow k = -\dfrac{5}{2}$.

Substituting into (v): $h = 11 – \dfrac{15}{2} = \dfrac{7}{2}$.

Finding $r^2$ from (ii):

$$r^2 = \left(2-\frac{7}{2}\right)^2 + \left(3+\frac{5}{2}\right)^2 = \frac{9}{4} + \frac{121}{4} = \frac{130}{4}$$

The equation of the required circle:

$$\left(x-\frac{7}{2}\right)^2+\left(y+\frac{5}{2}\right)^2=\frac{130}{4}$$ $$\left(x^2-7x+\frac{49}{4}\right)+\left(y^2+5y+\frac{25}{4}\right)=\frac{130}{4}$$ $$\boxed{x^2 + y^2 – 7x + 5y – 14 = 0}$$

Since the centre lies on the $x$-axis, its $y$-coordinate is 0. Let the centre be $C(h, 0)$ and radius $= 5$.

The equation of the circle becomes:

$$(x-h)^2 + y^2 = 25 \quad\cdots(i)$$

Since the circle passes through $(2,3)$, substituting $x = 2$, $y = 3$:

$$4 + 9 – 4h + h^2 – 25 = 0 \Rightarrow h^2 – 4h – 12 = 0 \Rightarrow (h-6)(h+2) = 0$$ $$\therefore\; h = 6 \quad\text{or}\quad h = -2$$

When $h = 6$: $\quad x^2 + y^2 – 12x + 11 = 0$   $(\because 36 – 25 = 11)$

When $h = -2$: $\quad x^2 + y^2 + 4x – 21 = 0$   $(\because 4 – 25 = -21)$

Since the circle passes through the origin $O(0,0)$ and makes intercepts $a$ and $b$, it also passes through $A(a,0)$ and $B(0,b)$.

Since $\angle AOB = 90°$, the chord $AB$ is a diameter (angle in a semicircle = 90°). The centre $C$ is the midpoint of $AB$:

$$C = \left(\frac{a}{2},\, \frac{b}{2}\right)$$

The radius is:

$$r = \frac{1}{2}\sqrt{a^2 + b^2}$$

Using the standard circle equation:

$$\left(x – \frac{a}{2}\right)^2 + \left(y – \frac{b}{2}\right)^2 = \frac{1}{4}(a^2 + b^2)$$ $$x^2 + \frac{a^2}{4} – ax + y^2 + \frac{b^2}{4} – by = \frac{a^2}{4} + \frac{b^2}{4}$$ $$\boxed{x^2 + y^2 – ax – by = 0}$$

The centre is $C(2,2)$. Since the circle passes through $P(4,5)$, the radius equals $CP$:

$$r = \sqrt{(4-2)^2 + (5-2)^2} = \sqrt{4+9} = \sqrt{13}$$

The equation of the circle is:

$$(x-2)^2 + (y-2)^2 = (\sqrt{13})^2$$ $$(x^2 – 4x + 4) + (y^2 – 4y + 4) = 13$$ $$\boxed{x^2 + y^2 – 4x – 4y = 5}$$

The circle $x^2 + y^2 = 5^2$ has its centre at $O(0,0)$ and radius $5$.

We compute the distance from the given point $P(-2.5,\, 3.5)$ to the centre $O$:

$$OP = \sqrt{(2.5)^2 + (3.5)^2} = \sqrt{6.25 + 12.25} = \sqrt{18.50} < \sqrt{25} = 5$$

Since $OP <$ radius, the point $P(-2.5,\, 3.5)$ lies inside the circle.

Comparing $y^2 = 12x$ with the standard form $y^2 = 4ax$ (Standard Form I):

$$4a = 12 \Rightarrow a = 3$$

• Focus: $(a, 0) = (3, 0)$
• Axis: $x$-axis
• Directrix: $x = -a$, i.e., $x = -3$
• Length of latus rectum: $4a = 12$

The given equation $x^2 = 6y$ matches Standard Form III $x^2 = 4ay$:

$$4a = 6 \Rightarrow a = \frac{3}{2}$$

• Focus: $(0, a) = \left(0, \dfrac{3}{2}\right)$
• Axis: $y$-axis
• Directrix: $y = -a$, i.e., $y = -\dfrac{3}{2}$
• Length of latus rectum: $4a = 6$

Comparing $y^2 = -8x$ with Standard Form II $y^2 = -4ax$:

$$4a = 8 \Rightarrow a = 2$$

• Focus: $(-a, 0) = (-2, 0)$
• Axis: $x$-axis
• Directrix: $x = a$, i.e., $x = 2$
• Length of latus rectum: $4a = 8$

The given equation $x^2 = -16y$ matches Standard Form IV $x^2 = -4ay$:

$$4a = 16 \Rightarrow a = 4$$

• Focus: $(0, -a) = (0, -4)$
• Axis: $y$-axis
• Directrix: $y = a$, i.e., $y = 4$
• Length of latus rectum: $4a = 16$

Comparing $y^2 = 10x$ with Standard Form I $y^2 = 4ax$:

$$4a = 10 \Rightarrow a = \frac{5}{2}$$

• Focus: $(a, 0) = \left(\dfrac{5}{2}, 0\right)$
• Axis: $x$-axis
• Directrix: $x = -a$, i.e., $x = -\dfrac{5}{2}$
• Length of latus rectum: $4a = 10$

Comparing $x^2 = -9y$ with Standard Form IV $x^2 = -4ay$:

$$4a = 9 \Rightarrow a = \frac{9}{4}$$

• Focus: $(0, -a) = \left(0, -\dfrac{9}{4}\right)$
• Axis: $y$-axis
• Directrix: $y = a$, i.e., $y = \dfrac{9}{4}$
• Length of latus rectum: $4a = 9$

The focus $(6,0)$ lies on the $x$-axis, so the axis of the parabola is the $x$-axis. Since the focus lies to the right of the vertex (origin), the parabola opens rightward and has the form $y^2 = 4ax$.

Comparing focus $(a, 0) = (6, 0)$: $\quad a = 6$.

$$\boxed{y^2 = 24x}$$

The focus $(0,-3)$ lies on the $y$-axis, so the axis of the parabola is the $y$-axis. Since the focus lies below the origin and the directrix $y = 3$ is above, the parabola opens downward: $x^2 = -4ay$ with $a = 3$.

$$x^2 = -4(3)y$$ $$\boxed{x^2 = -12y}$$

The focus $(3,0)$ lies on the $x$-axis to the right of the origin, so the parabola opens rightward: $y^2 = 4ax$ with $a = 3$.

$$\boxed{y^2 = 12x}$$

The focus $(-2,0)$ lies on the $x$-axis to the left of the origin, so the parabola opens leftward: $y^2 = -4ax$ with $a = 2$.

$$y^2 = -4(2)x$$ $$\boxed{y^2 = -8x}$$

Since the axis is along the $x$-axis and the parabola passes through $(2,3)$ which is in the first quadrant, it must open rightward. So the equation is of the form:

$$y^2 = 4ax \quad\cdots(i)$$

Substituting the point $(2,3)$:

$$3^2 = 4a(2) \Rightarrow a = \frac{9}{8}$$

Substituting $a$ back into (i):

$$y^2 = 4\left(\frac{9}{8}\right)x = \frac{9}{2}x$$ $$\boxed{2y^2 = 9x}$$

Since the parabola is symmetric about the $y$-axis and passes through $(5,2)$ in the first quadrant, it opens upward: $x^2 = 4ay$  …(i)

Substituting $(5,2)$:

$$5^2 = 4a(2) \Rightarrow 25 = 8a \Rightarrow a = \frac{25}{8}$$

Substituting $a$ back:

$$x^2 = 4\left(\frac{25}{8}\right)y = \frac{25}{2}y$$ $$\boxed{2x^2 = 25y}$$

Since $36 > 16$, the major axis is along the $x$-axis. Comparing with $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$: $a^2 = 36$, $b^2 = 16$, so $a = 6$, $b = 4$.

$$c = \sqrt{a^2-b^2} = \sqrt{36-16} = \sqrt{20} = 2\sqrt{5}$$

• Foci: $(\pm 2\sqrt{5},\, 0)$
• Vertices: $(\pm 6,\, 0)$
• Major axis length: $2a = 12$
• Minor axis length: $2b = 8$
• Eccentricity: $e = \dfrac{c}{a} = \dfrac{2\sqrt{5}}{6} = \dfrac{\sqrt{5}}{3}$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{2\times16}{6} = \dfrac{16}{3}$

Since $25 > 4$, the major axis is along the $y$-axis. Comparing with $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$: $a^2 = 25$, $b^2 = 4$, so $a = 5$, $b = 2$.

$$c = \sqrt{a^2-b^2} = \sqrt{25-4} = \sqrt{21}$$

• Foci: $(0,\, \pm\sqrt{21})$
• Vertices: $(0,\, \pm 5)$
• Major axis length: $2a = 10$
• Minor axis length: $2b = 4$
• Eccentricity: $e = \dfrac{\sqrt{21}}{5}$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{8}{5}$

Since $16 > 9$, the major axis is along the $x$-axis. We have $a^2 = 16$, $b^2 = 9$, so $a = 4$, $b = 3$.

$$c = \sqrt{16-9} = \sqrt{7}$$

• Foci: $(\pm\sqrt{7},\, 0)$
• Vertices: $(\pm 4,\, 0)$
• Major axis length: $2a = 8$
• Minor axis length: $2b = 6$
• Eccentricity: $e = \dfrac{\sqrt{7}}{4}$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{9}{2}$

Since $100 > 25$, the major axis is along the $y$-axis. We have $a^2 = 100$, $b^2 = 25$, so $a = 10$, $b = 5$.

$$c = \sqrt{100-25} = \sqrt{75} = 5\sqrt{3}$$

• Foci: $(0,\, \pm 5\sqrt{3})$
• Vertices: $(0,\, \pm 10)$
• Major axis length: $2a = 20$
• Minor axis length: $2b = 10$
• Eccentricity: $e = \dfrac{5\sqrt{3}}{10} = \dfrac{\sqrt{3}}{2}$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{2\times25}{10} = 5$

Since $49 > 36$, the major axis is along the $x$-axis. We have $a^2 = 49$, $b^2 = 36$, so $a = 7$, $b = 6$.

$$c = \sqrt{49-36} = \sqrt{13}$$

• Foci: $(\pm\sqrt{13},\, 0)$
• Vertices: $(\pm 7,\, 0)$
• Major axis length: $2a = 14$
• Minor axis length: $2b = 12$
• Eccentricity: $e = \dfrac{\sqrt{13}}{7}$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{72}{7}$

Since $400 > 100$, the major axis is along the $y$-axis. We have $a^2 = 400$, $b^2 = 100$, so $a = 20$, $b = 10$.

$$c = \sqrt{400-100} = \sqrt{300} = 10\sqrt{3}$$

• Foci: $(0,\, \pm 10\sqrt{3})$
• Vertices: $(0,\, \pm 20)$
• Major axis length: $2a = 40$
• Minor axis length: $2b = 20$
• Eccentricity: $e = \dfrac{10\sqrt{3}}{20} = \dfrac{\sqrt{3}}{2}$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{200}{20} = 10$

Dividing both sides by 144 to get the standard form:

$$\frac{x^2}{4} + \frac{y^2}{36} = 1$$

Since $36 > 4$, the major axis is along the $y$-axis. We have $a^2 = 36$, $b^2 = 4$, so $a = 6$, $b = 2$.

$$c = \sqrt{36-4} = \sqrt{32} = 4\sqrt{2}$$

• Foci: $(0,\, \pm 4\sqrt{2})$
• Vertices: $(0,\, \pm 6)$
• Major axis length: $2a = 12$
• Minor axis length: $2b = 4$
• Eccentricity: $e = \dfrac{4\sqrt{2}}{6} = \dfrac{2\sqrt{2}}{3}$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{8}{6} = \dfrac{4}{3}$

Dividing by 16:

$$\frac{x^2}{1} + \frac{y^2}{16} = 1$$

Since $16 > 1$, the major axis is along the $y$-axis. We have $a^2 = 16$, $b^2 = 1$, so $a = 4$, $b = 1$.

$$c = \sqrt{16-1} = \sqrt{15}$$

• Foci: $(0,\, \pm\sqrt{15})$
• Vertices: $(0,\, \pm 4)$
• Major axis length: $2a = 8$
• Minor axis length: $2b = 2$
• Eccentricity: $e = \dfrac{\sqrt{15}}{4}$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{2}{4} = \dfrac{1}{2}$

Dividing by 36:

$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$

Since $9 > 4$, the major axis is along the $x$-axis. We have $a^2 = 9$, $b^2 = 4$, so $a = 3$, $b = 2$.

$$c = \sqrt{9-4} = \sqrt{5}$$

• Foci: $(\pm\sqrt{5},\, 0)$
• Vertices: $(\pm 3,\, 0)$
• Major axis length: $2a = 6$
• Minor axis length: $2b = 4$
• Eccentricity: $e = \dfrac{\sqrt{5}}{3}$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{8}{3}$

Since the vertices and foci lie on the $x$-axis, the major axis is along the $x$-axis. The ellipse equation is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$.

From vertices: $a = 5 \Rightarrow a^2 = 25$. From foci: $c = 4$.

$$c^2 = a^2 – b^2 \Rightarrow 16 = 25 – b^2 \Rightarrow b^2 = 9$$ $$\boxed{\frac{x^2}{25} + \frac{y^2}{9} = 1}$$

Since the vertices and foci lie on the $y$-axis, the major axis is along the $y$-axis. The ellipse equation is $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$.

From vertices: $a = 13$. From foci: $c = 5$.

$$c^2 = a^2 – b^2 \Rightarrow 25 = 169 – b^2 \Rightarrow b^2 = 144$$ $$\boxed{\frac{x^2}{144} + \frac{y^2}{169} = 1}$$

Major axis is along the $x$-axis. From vertices: $a = 6$. From foci: $c = 4$.

$$c^2 = a^2 – b^2 \Rightarrow 16 = 36 – b^2 \Rightarrow b^2 = 20$$ $$\boxed{\frac{x^2}{36} + \frac{y^2}{20} = 1}$$

Major axis is along the $x$-axis. From ends of major axis: $a = 3$. From ends of minor axis: $b = 2$.

$$\boxed{\frac{x^2}{9} + \frac{y^2}{4} = 1}$$

Major axis is along the $y$-axis. From ends of major axis: $a = \sqrt{5}$. From ends of minor axis: $b = 1$.

$$\boxed{\frac{x^2}{1} + \frac{y^2}{5} = 1}$$

Foci on $x$-axis, so major axis along $x$-axis. $2a = 26 \Rightarrow a = 13$. From foci: $c = 5$.

$$b^2 = a^2 – c^2 = 169 – 25 = 144$$ $$\boxed{\frac{x^2}{169} + \frac{y^2}{144} = 1}$$

Foci on $y$-axis, so major axis along $y$-axis. $2b = 16 \Rightarrow b = 8$. From foci: $c = 6$.

$$a^2 = b^2 + c^2 = 64 + 36 = 100$$ $$\boxed{\frac{x^2}{64} + \frac{y^2}{100} = 1}$$

Foci on $x$-axis. From foci: $c = 3$. Given $a = 4$.

$$b^2 = a^2 – c^2 = 16 – 9 = 7$$ $$\boxed{\frac{x^2}{16} + \frac{y^2}{7} = 1}$$

With foci on the $x$-axis and centre at the origin, the ellipse takes the form $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$.

Using the relation $a^2 = b^2 + c^2$:

$$a^2 = 3^2 + 4^2 = 9 + 16 = 25$$ $$\boxed{\frac{x^2}{25} + \frac{y^2}{9} = 1}$$

With major axis on the $y$-axis, the ellipse is $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$  …(i)

Substituting $(3,2)$: $\dfrac{9}{b^2} + \dfrac{4}{a^2} = 1$  …(ii)

Substituting $(1,6)$: $\dfrac{1}{b^2} + \dfrac{36}{a^2} = 1$  …(iii)

Multiplying (iii) by 9: $\dfrac{9}{b^2} + \dfrac{324}{a^2} = 9$  …(iv)

Subtracting (ii) from (iv):

$$\frac{320}{a^2} = 8 \Rightarrow a^2 = 40$$

Substituting $a^2 = 40$ into (iii):

$$\frac{1}{b^2} + \frac{36}{40} = 1 \Rightarrow \frac{1}{b^2} = \frac{1}{10} \Rightarrow b^2 = 10$$ $$\boxed{\frac{x^2}{10} + \frac{y^2}{40} = 1}$$

The ellipse equation is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$  …(i)

Substituting $(4,3)$: $\dfrac{16}{a^2} + \dfrac{9}{b^2} = 1$  …(ii)

Substituting $(6,2)$: $\dfrac{36}{a^2} + \dfrac{4}{b^2} = 1$  …(iii)

Multiplying (ii) by 4 and (iii) by 9, then subtracting to eliminate $b^2$:

$$\frac{64-324}{a^2} = 4-9 \Rightarrow \frac{-260}{a^2} = -5 \Rightarrow a^2 = 52$$

Substituting $a^2 = 52$ into (ii):

$$\frac{16}{52} + \frac{9}{b^2} = 1 \Rightarrow \frac{9}{b^2} = \frac{9}{13} \Rightarrow b^2 = 13$$ $$\frac{x^2}{52} + \frac{y^2}{13} = 1 \quad\Rightarrow\quad \boxed{x^2 + 4y^2 = 52}$$

Comparing with $\dfrac{x^2}{a^2} – \dfrac{y^2}{b^2} = 1$: $a^2 = 16$, $b^2 = 9$, so $a = 4$, $b = 3$.

$$c = \sqrt{a^2+b^2} = \sqrt{16+9} = 5$$

• Foci: $(\pm 5,\, 0)$
• Vertices: $(\pm 4,\, 0)$
• Eccentricity: $e = \dfrac{c}{a} = \dfrac{5}{4}$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{18}{4} = \dfrac{9}{2}$

Comparing with $\dfrac{y^2}{a^2} – \dfrac{x^2}{b^2} = 1$: $a^2 = 9$, $b^2 = 27$, so $a = 3$, $b = 3\sqrt{3}$.

$$c = \sqrt{9+27} = \sqrt{36} = 6$$

• Foci: $(0,\, \pm 6)$
• Vertices: $(0,\, \pm 3)$
• Eccentricity: $e = \dfrac{6}{3} = 2$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{54}{3} = 18$

Dividing both sides by 36 to get standard form:

$$\frac{y^2}{4} – \frac{x^2}{9} = 1$$

Comparing with $\dfrac{y^2}{a^2} – \dfrac{x^2}{b^2} = 1$: $a^2 = 4$, $b^2 = 9$, so $a = 2$, $b = 3$.

$$c = \sqrt{4+9} = \sqrt{13}$$

• Foci: $(0,\, \pm\sqrt{13})$
• Vertices: $(0,\, \pm 2)$
• Eccentricity: $e = \dfrac{\sqrt{13}}{2}$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{18}{2} = 9$

Dividing by 576:

$$\frac{x^2}{36} – \frac{y^2}{64} = 1$$

We have $a^2 = 36$, $b^2 = 64$, so $a = 6$, $b = 8$.

$$c = \sqrt{36+64} = \sqrt{100} = 10$$

• Foci: $(\pm 10,\, 0)$
• Vertices: $(\pm 6,\, 0)$
• Eccentricity: $e = \dfrac{10}{6} = \dfrac{5}{3}$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{128}{6} = \dfrac{64}{3}$

Dividing by 36:

$$\frac{y^2}{36/5} – \frac{x^2}{4} = 1$$

We have $a^2 = \dfrac{36}{5}$, $b^2 = 4$, so $a = \dfrac{6}{\sqrt{5}}$, $b = 2$.

$$c = \sqrt{\frac{36}{5}+4} = \sqrt{\frac{56}{5}} = \frac{2\sqrt{14}}{\sqrt{5}}$$

• Foci: $\left(0,\, \pm\dfrac{2\sqrt{14}}{\sqrt{5}}\right)$
• Vertices: $\left(0,\, \pm\dfrac{6}{\sqrt{5}}\right)$
• Eccentricity: $e = \dfrac{\sqrt{14}}{3}$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{4\sqrt{5}}{3}$

Dividing by 784:

$$\frac{y^2}{16} – \frac{x^2}{49} = 1$$

We have $a^2 = 16$, $b^2 = 49$, so $a = 4$, $b = 7$.

$$c = \sqrt{16+49} = \sqrt{65}$$

• Foci: $(0,\, \pm\sqrt{65})$
• Vertices: $(0,\, \pm 4)$
• Eccentricity: $e = \dfrac{\sqrt{65}}{4}$
• Latus rectum: $\dfrac{2b^2}{a} = \dfrac{98}{4} = \dfrac{49}{2}$

Foci on $x$-axis, so the hyperbola is $\dfrac{x^2}{a^2} – \dfrac{y^2}{b^2} = 1$.

From vertices: $a = 2$. From foci: $c = 3$.

$$b^2 = c^2 – a^2 = 9 – 4 = 5$$ $$\boxed{\frac{x^2}{4} – \frac{y^2}{5} = 1}$$

Foci on $y$-axis, so the hyperbola is $\dfrac{y^2}{a^2} – \dfrac{x^2}{b^2} = 1$.

From vertices: $a = 5$. From foci: $c = 8$.

$$b^2 = c^2 – a^2 = 64 – 25 = 39$$ $$\boxed{\frac{y^2}{25} – \frac{x^2}{39} = 1}$$

Foci on $y$-axis, so the hyperbola is $\dfrac{y^2}{a^2} – \dfrac{x^2}{b^2} = 1$.

From vertices: $a = 3$. From foci: $c = 5$.

$$b^2 = c^2 – a^2 = 25 – 9 = 16$$ $$\boxed{\frac{y^2}{9} – \frac{x^2}{16} = 1}$$

Foci on $x$-axis: $c = 5$. Transverse axis $2a = 8 \Rightarrow a = 4$, $a^2 = 16$.

$$b^2 = c^2 – a^2 = 25 – 16 = 9$$ $$\boxed{\frac{x^2}{16} – \frac{y^2}{9} = 1}$$

Foci on $y$-axis: $c = 13$. Conjugate axis $2b = 24 \Rightarrow b = 12$, $b^2 = 144$.

$$a^2 = c^2 – b^2 = 169 – 144 = 25$$ $$\boxed{\frac{y^2}{25} – \frac{x^2}{144} = 1}$$

Foci on $x$-axis: $c = 3\sqrt{5}$. From the latus rectum condition: $\dfrac{2b^2}{a} = 8 \Rightarrow b^2 = 4a$  …(ii)

Using $c^2 = a^2 + b^2$: $45 = a^2 + 4a \Rightarrow a^2 + 4a – 45 = 0 \Rightarrow (a+9)(a-5) = 0$.

Since $a > 0$: $a = 5$, $a^2 = 25$, and $b^2 = 4 \times 5 = 20$.

$$\boxed{\frac{x^2}{25} – \frac{y^2}{20} = 1}$$

Foci on $x$-axis: $c = 4$. From the latus rectum: $\dfrac{2b^2}{a} = 12 \Rightarrow b^2 = 6a$  …(ii)

Using $c^2 = a^2 + b^2$: $16 = a^2 + 6a \Rightarrow a^2 + 6a – 16 = 0 \Rightarrow (a+8)(a-2) = 0$.

Since $a > 0$: $a = 2$, $a^2 = 4$, and $b^2 = 6 \times 2 = 12$.

$$\boxed{\frac{x^2}{4} – \frac{y^2}{12} = 1}$$

Vertices on $x$-axis: $a = 7$, $a^2 = 49$.

$$e = \frac{c}{a} = \frac{4}{3} \Rightarrow c = \frac{4 \times 7}{3} = \frac{28}{3}$$ $$b^2 = c^2 – a^2 = \frac{784}{9} – 49 = \frac{784 – 441}{9} = \frac{343}{9}$$ $$\boxed{\frac{x^2}{49} – \frac{9y^2}{343} = 1}$$

Foci on $y$-axis: $c = \sqrt{10}$. The hyperbola is $\dfrac{y^2}{a^2} – \dfrac{x^2}{b^2} = 1$ with $b^2 = c^2 – a^2 = 10 – a^2$.

Since $(2,3)$ lies on it:

$$\frac{9}{a^2} – \frac{4}{10-a^2} = 1$$ $$9(10-a^2) – 4a^2 = a^2(10-a^2)$$ $$\Rightarrow a^4 – 23a^2 + 90 = 0$$

Let $t = a^2$: $t^2 – 23t + 90 = 0 \Rightarrow (t-18)(t-5) = 0 \Rightarrow t = 18$ or $t = 5$.

When $a^2 = 18$: $b^2 = 10 – 18 = -8 < 0$ (rejected). So $a^2 = 5$, $b^2 = 10 - 5 = 5$.

$$\frac{y^2}{5} – \frac{x^2}{5} = 1 \quad\Rightarrow\quad \boxed{y^2 – x^2 = 5}$$

We place the vertex of the parabola at the origin with the axis along the positive $x$-axis. The equation of the parabolic section is a rightward parabola:

$$y^2 = 4ax \quad\cdots(i)$$

The depth $OA = 5$ cm and diameter $LM = 20$ cm, so $AL = AM = 10$ cm.

Therefore, the coordinates of point $L$ are $(5,\, 10)$. Since $L$ lies on parabola (i):

$$(10)^2 = 4a(5) \Rightarrow 100 = 20a \Rightarrow a = 5$$

∴ Focus is at $(a, 0) = (5, 0)$, which is exactly at point $A$ — the midpoint of the diameter.

We place the vertex at the origin $O$ with axis along $OY’$ (downward). Since it’s a downward-opening arch, the equation is:

$$x^2 = -4ay \quad\cdots(i)$$

The half-width at the base is $MB = \dfrac{5}{2} = 2.5$ m and depth $OM = 10$ m, so point $B = (2.5,\, -10)$.

Since $B$ lies on (i): $(2.5)^2 = -4a(-10) \Rightarrow 6.25 = 40a \Rightarrow a = \dfrac{6.25}{40}$

Substituting $a$ back into (i):

$$x^2 = -\frac{6.25}{10}y = -0.625y \quad\cdots(ii)$$

At a depth of 2 m from the vertex, the point $D = (x,\, -2)$ lies on (ii):

$$x^2 = -0.625 \times (-2) = 1.25 = \frac{5}{4} \Rightarrow x = \frac{\sqrt{5}}{2}$$

Width $CD = 2x = 2 \times \dfrac{\sqrt{5}}{2} = \sqrt{5} \approx \mathbf{2.33\; m}$.

The cable forms an upward parabola with vertex at the lowest point $O$. The equation is $x^2 = 4ay$  …(i).

Since $LM = 100$ m, we have $AL = AM = 50$ m. Point $B$ has coordinates: $x = OH = 50$ and $y = BH = BM – OA = 30 – 6 = 24$.

Substituting into (i): $(50)^2 = 4a(24) \Rightarrow a = \dfrac{2500}{96} = \dfrac{625}{24}$.

The equation becomes:

$$x^2 = \frac{625}{6}y \quad\cdots(ii)$$

At $AN = 18$ m, the wire $NP$ meets the parabola at point $P$ with $x = OK = 18$ and $y = PK = PN – OA = PN – 6$.

Substituting into (ii): $(18)^2 = \dfrac{625}{6}(PN – 6) \Rightarrow 324 \times 6 = 625(PN – 6)$

$$PN – 6 = \frac{1944}{625} = 3.2704 \Rightarrow PN = 6 + 3.2704 = \mathbf{9.2704\; m}$$

The arch is a semi-ellipse with semi-major axis $a = \dfrac{8}{2} = 4$ m and semi-minor axis $b = OB = 2$ m.

The equation of the ellipse is:

$$\frac{x^2}{16} + \frac{y^2}{4} = 1 \quad\cdots(i)$$

We need the height at a point $1.5$ m from one end (say $A$). Since $OA = 4$:

$$OP = OA – AP = 4 – 1.5 = 2.5\text{ m}$$

At point $R$, $x = OP = \dfrac{5}{2}$. Substituting into (i):

$$\frac{25/4}{16} + \frac{y^2}{4} = 1 \Rightarrow \frac{y^2}{4} = 1 – \frac{25}{64} = \frac{39}{64} \Rightarrow y^2 = \frac{39}{16}$$ $$y = \frac{\sqrt{39}}{4} \approx \frac{6.24}{4} \approx \mathbf{1.56\; m}$$

Let $AB$ be the rod of length 12 cm with $A$ on the $x$-axis and $B$ on the $y$-axis. Point $P(x,y)$ is on the rod with $AP = 3$ cm, so $PB = 12 – 3 = 9$ cm.

Draw $PL \perp y$-axis and $PM \perp x$-axis, so $PL = x$ and $PM = y$.

Let $\angle OAB = \theta$, so $\angle LPB = \theta$ (corresponding angles).

In right-angled $\triangle AMP$: $\sin\theta = \dfrac{MP}{AP} = \dfrac{y}{3}$  …(i)

In right-angled $\triangle PLB$: $\cos\theta = \dfrac{PL}{PB} = \dfrac{x}{9}$  …(ii)

Using the identity $\cos^2\theta + \sin^2\theta = 1$:

$$\left(\frac{x}{9}\right)^2 + \left(\frac{y}{3}\right)^2 = 1$$ $$\boxed{\frac{x^2}{81} + \frac{y^2}{9} = 1}$$

This is an ellipse.

Comparing $x^2 = 12y$ with $x^2 = 4ay$: $4a = 12 \Rightarrow a = 3$.

The latus rectum $AB$ has length $4a = 12$. The focus $S$ is at $(0, a) = (0, 3)$, so the distance $OS = 3$.

The triangle $OAB$ has base $AB = 12$ and height $OS = 3$:

$$\text{Area of } \triangle OAB = \frac{1}{2} \times AB \times OS = \frac{1}{2} \times 12 \times 3 = \mathbf{18 \text{ sq. units}}$$

Let $S$ and $S’$ be the two flag posts. Since $PS + PS’ = 10$ m (constant), by the definition of an ellipse, the path is an ellipse with foci at the two flag posts.

Length of major axis $= 2a = 10 \Rightarrow a = 5$.

Distance between foci $= 2c = 8 \Rightarrow c = 4$.

Using $c^2 = a^2 – b^2$:

$$16 = 25 – b^2 \Rightarrow b^2 = 9$$ $$\boxed{\frac{x^2}{25} + \frac{y^2}{9} = 1}$$

The parabola is $y^2 = 4ax$ with vertex at the origin $O$. Let $OAB$ be the equilateral triangle with each side of length $l$.

By symmetry, $AB \perp OX$ and $\angle BOC = \angle AOC = 30°$.

In $\triangle OCA$:

$$\cos 30° = \frac{x}{l} \Rightarrow x = \frac{l\sqrt{3}}{2}$$ $$\sin 30° = \frac{y}{l} \Rightarrow y = \frac{l}{2}$$

So the coordinates of $A$ are $\left(\dfrac{l\sqrt{3}}{2},\, \dfrac{l}{2}\right)$. Since $A$ lies on the parabola:

$$\left(\frac{l}{2}\right)^2 = 4a \cdot \frac{l\sqrt{3}}{2} \Rightarrow \frac{l^2}{4} = 2a\sqrt{3}\,l$$ $$\Rightarrow l = 8\sqrt{3}\,a$$

∴ The length of each side of the equilateral triangle is $\mathbf{8\sqrt{3}\,a}$.