1. Discuss each of the following conic sections and sketch the graph: (a) $$ 63 y^{2}+42 y-18 x^{2}+36 x \sqrt{2}=155 $$ (b) $$ 25 y^{2}+100 y-4 x^{2}-48 x-44=0 $$ 2. A conic section with centre at $$ (\sqrt{2},-1) $$ has the value of eccentricity given by $$ e=\frac{3 \sqrt{2}}{2} $$. Given that the focal axis is horizontal and that the distance between foci and directrices is seven thirds, (a) write down the conic section's equation (b) write down the equations of directrices. 3. The opening of a tunnel is in the shape of a semi-elliptical arch. The arch is 18 metres wide and 12 metres high. Find the height of the arch 4 metres from the edge of the tunnel.

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Let's solve each part of the problem step by step.

### 1. Discuss and sketch the graphs of the conic sections.

#### (a) $$ 63 y^{2}+42 y-18 x^{2}+36 x \sqrt{2}=155 $$

1. **Rearranging the equation:**
   $$
   63 y^{2} + 42 y - 18 x^{2} + 36 x \sqrt{2} - 155 = 0
   $$

2. **Identifying the conic section:**
   - The equation has both $$ y^{2} $$ and $$ x^{2} $$ terms, indicating it could be a hyperbola or an ellipse. 
   - To determine the type, we can rewrite it in standard form.

3. **Completing the square for $$ y $$ and $$ x $$:**
   - For $$ y $$:
     $$
     63(y^{2} + \frac{42}{63}y) = 63(y^{2} + \frac{2}{3}y) = 63\left(y + \frac{1}{3}\right)^{2} - 7
     $$
   - For $$ x $$:
     $$
     -18(x^{2} - 2\sqrt{2}x) = -18\left((x - \sqrt{2})^{2} - 2\right) = -18(x - \sqrt{2})^{2} + 36
     $$

4. **Substituting back:**
   $$
   63\left(y + \frac{1}{3}\right)^{2} - 7 - 18(x - \sqrt{2})^{2} + 36 - 155 = 0
   $$
   $$
   63\left(y + \frac{1}{3}\right)^{2} - 18(x - \sqrt{2})^{2} - 126 = 0
   $$
   $$
   \frac{63\left(y + \frac{1}{3}\right)^{2}}{126} - \frac{18(x - \sqrt{2})^{2}}{126} = 1
   $$
   $$
   \frac{\left(y + \frac{1}{3}\right)^{2}}{2} - \frac{(x - \sqrt{2})^{2}}{7} = 1
   $$

5. **Conclusion:**
   - This is a hyperbola centered at $$ (\sqrt{2}, -\frac{1}{3}) $$ with $$ a^2 = 2 $$ and $$ b^2 = 7 $$.

#### (b) $$ 25 y^{2}+100 y-4 x^{2}-48 x-44=0 $$

1. **Rearranging the equation:**
   $$
   25 y^{2} + 100 y - 4 x^{2} - 48 x - 44 = 0
   $$

2. **Identifying the conic section:**
   - Similar to part (a), we will complete the square.

3. **Completing the square for $$ y $$ and $$ x $$:**
   - For $$ y $$:
     $$
     25(y^{2} + 4y) = 25((y + 2)^{2} - 4) = 25(y + 2)^{2} - 100
     $$
   - For $$ x $$:
     $$
     -4(x^{2} + 12x) = -4((x + 6)^{2} - 36) = -4(x + 6)^{2} + 144
     $$

4. **Substituting back:**
   $$
   25(y + 2)^{2} - 100 - 4(x + 6)^{2} + 144 - 44 = 0
   $$
   $$
   25(y + 2)^{2} - 4(x + 6)^{2} = 0
   $$

5. **Conclusion:**
   - This is a degenerate conic (a pair of lines) since it can be factored.

### 2. Conic section with given properties.

#### (a) Write down the conic section's equation.

1. **Given properties:**
   - Center: $$ (\sqrt{2}, -1) $$
   - Eccentricity: $$ e = \frac{3\sqrt{2}}{2} $$
   - Distance between foci and directrices: $$ \frac{7}{3} $$

2. **Finding $$ a $$ and $$ b $$:**
   - The distance between foci $$ 2c = 7 $$ implies $$ c = \frac{7}{2} $$.
   - Using $$ c = ae $$, we have:
     $$
     \frac{7}{2} = a \cdot \frac{3\sqrt{2}}{2} \implies a = \frac{7}{3\sqrt{2}} = \frac{7\sqrt{2}}{6}
     $$
   - Now, $$ b = \sqrt{c^2 - a^2} $$:
     $$
     c^2 = \left(\frac{7}{2}\right)^2 = \frac{49}{4}, \quad a^2 = \left(\frac{7\sqrt{2}}{6}\right)^2 = \frac{49 \cdot 2}{36} = \frac{98}{36} = \frac{49}{18}
     $$
     $$
     b^2 = \frac{49}{4} - \frac{49}{18} = \frac{49 \cdot 9 - 49 \cdot 4}{36} = \frac{49(9 - 4)}{36} = \frac{49 \cdot 5}{36} = \frac{245}{36}
     $$

3. **Equation of the hyperbola:**
   $$
   \frac{(x - \sqrt{2})^2}{\left(\frac{7\sqrt{2}}{6}\right)^2} - \frac{(y + 1)^2}{\frac{245}{36}} = 1
   $$

#### (b) Write down the equations of directrices.

1. **Directrices for a hyperbola:**
   - The directrices are given by $$ x = h \pm \frac{a}{e} $$:
   $$
   \frac{7\sqrt{2}}{6} \div \frac{3\sqrt{2}}{2} = \frac{7}{9}
   $$
   - Thus, the equations of the directrices are:
   $$
   x = \sqrt{2} \pm \frac{7}{9}

1. **Conic Sections:** - **(a)** The equation $$63 y^{2}+42 y-18 x^{2}+36 x \sqrt{2}=155$$ represents a hyperbola centered at $$ (\sqrt{2}, -\frac{1}{3}) $$ with $$ a^2 = 2 $$ and $$ b^2 = 7 $$. - **(b)** The equation $$25 y^{2}+100 y-4 x^{2}-48 x-44=0$$ is a degenerate conic, specifically a pair of lines. 2. **Conic Section with Given Properties:** - **(a)** The equation of the hyperbola is: \[ \frac{(x - \sqrt{2})^2}{\left(\frac{7\sqrt{2}}{6}\right)^2} - \frac{(y + 1)^2}{\frac{245}{36}} = 1 $$ - **(b)** The equations of the directrices are: $$ x = \sqrt{2} \pm \frac{7}{9} $$ 3. **Semi-Elliptical Arch Tunnel:** - The height of the arch 4 metres from the edge is 10 metres.

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