1. Find the foci, vertices and sketch each of the following ellipses: $$ \begin{array}{ll} ext { (a) } \frac{x^{2}}{25}+\frac{y^{2}}{9}=1 & ext { (b) } 4 x^{2}+y^{2}=1 .\end{array} $$ 2. Find the equation of the ellipse centred at the origin with vertical focal axis if semimajor aris is 5 and the distance between the foci and vertices is 1 unit.

Question

1. Find the foci, vertices and sketch each of the following ellipses: $$ \begin{array}{ll}	ext { (a) } \frac{x^{2}}{25}+\frac{y^{2}}{9}=1 & 	ext { (b) } 4 x^{2}+y^{2}=1 .\end{array} $$ 2. Find the equation of the ellipse centred at the origin with vertical focal axis if semimajor aris is 5 and the distance between the foci and vertices is 1 unit.

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**1. (a) Ellipse:** $$ \frac{x^2}{25}+\frac{y^2}{9}=1 $$

- **Center:** The center is at the origin $$ (0,0) $$.

- **Semiaxes:**  
  Since the denominator under $$ x^2 $$ is bigger,  
  $$
  a^2 = 25 \quad \text{and} \quad b^2 = 9,
  $$
  so the semimajor axis is $$ a=5 $$ and the semiminor axis is $$ b=3 $$.

- **Vertices:**  
  The vertices lie along the $$ x $$-axis, at  
  $$
  (\pm a,\,0) = (\pm5,\,0).
  $$

- **Foci:**  
  The focal distance $$ c $$ is given by  
  $$
  c^2 = a^2 - b^2 = 25 - 9 = 16 \quad \Longrightarrow \quad c=4.
  $$
  Thus, the foci are at  
  $$
  (\pm c,\,0) = (\pm4,\,0).
  $$

---

**1. (b) Ellipse:** $$ 4x^2+y^2=1 $$

- **Rewrite in standard form:**  
  Divide by 1 to write  
  $$
  \frac{x^2}{\frac{1}{4}}+\frac{y^2}{1}=1.
  $$
  In this form, since $$ 1 > \frac{1}{4} $$, the major axis is vertical.

- **Semiaxes:**  
  Here,  
  $$
  a^2 = 1 \quad \text{and} \quad b^2 = \frac{1}{4},
  $$
  so $$ a=1 $$ and $$ b=\frac{1}{2} $$.

- **Vertices:**  
  With a vertical major axis, the vertices are at  
  $$
  (0, \pm a) = (0, \pm1).
  $$

- **Foci:**  
  The focal distance $$ c $$ is found using  
  $$
  c^2 = a^2 - b^2 = 1 - \frac{1}{4}=\frac{3}{4} \quad \Longrightarrow \quad c=\frac{\sqrt{3}}{2}.
  $$
  So the foci are at  
  $$
  (0, \pm c) = \left(0, \pm \frac{\sqrt{3}}{2}\right).
  $$

---

**2. Ellipse with vertical focal axis, semimajor axis $$ a=5 $$, and distance between a vertex and the corresponding focus equal to 1**

Since the ellipse is centered at the origin with a vertical major axis:
- **Vertices:** They are at  
  $$
  (0, \pm a) = (0, \pm5).
  $$

- **Distance from vertex to focus:**  
  The distance between a vertex $$ (0, \pm5) $$ and the corresponding focus $$ (0, \pm c) $$ is  
  $$
  a-c = 1.
  $$
  Thus,  
  $$
  c=5-1=4.
  $$

- **Find $$ b $$:**  
  For an ellipse with a vertical major axis,  
  $$
  c^2 = a^2 - b^2.
  $$
  Substituting,  
  $$
  4^2 = 5^2 - b^2 \quad \Longrightarrow \quad 16 = 25 - b^2 \quad \Longrightarrow \quad b^2 = 25 - 16 = 9.
  $$
  So, $$ b=3 $$.

- **Equation of the ellipse:**  
  The standard form for a vertical ellipse is  
  $$
  \frac{x^2}{b^2}+\frac{y^2}{a^2}=1,
  $$
  hence,  
  $$
  \frac{x^2}{9}+\frac{y^2}{25}=1.
  $$
**1. (a) Ellipse:** $$ \frac{x^2}{25}+\frac{y^2}{9}=1 $$ - **Center:** $$ (0,0) $$ - **Semiaxes:** $$ a=5 $$ (major axis), $$ b=3 $$ (minor axis) - **Vertices:** $$ (\pm5,0) $$ - **Foci:** $$ (\pm4,0) $$ --- **1. (b) Ellipse:** $$ 4x^2+y^2=1 $$ - **Center:** $$ (0,0) $$ - **Semiaxes:** $$ a=1 $$ (major axis), $$ b=0.5 $$ (minor axis) - **Vertices:** $$ (0,\pm1) $$ - **Foci:** $$ \left(0,\pm\frac{\sqrt{3}}{2}\right) $$ --- **2. Ellipse with vertical major axis, $$ a=5 $$, and $$ a-c=1 $$:** - **Center:** $$ (0,0) $$ - **Semiaxes:** $$ a=5 $$, $$ c=4 $$, $$ b=3 $$ - **Equation:** $$ \frac{x^2}{9}+\frac{y^2}{25}=1 $$

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