Find the distance between a focus and an extremity of the minor axis of the ellipse
(i) `4x^(2)+5y^(2)=100" (ii) "x^(2)/a^(2)+y^(2)/b^(2)=1`.

Answer

To find the distance between a focus and an extremity of the minor axis of the ellipse given by the equations, we will solve each part step by step.

### Part (i): For the ellipse \(4x^2 + 5y^2 = 100\)

**Step 1: Convert to standard form**
We start with the equation:
\[ 4x^2 + 5y^2 = 100 \]
Dividing the entire equation by 100, we get:
\[ \frac{x^2}{25} + \frac{y^2}{20} = 1 \]
This is now in the standard form of an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a^2 = 25\) and \(b^2 = 20\).

**Step 2: Identify values of \(a\) and \(b\)**
From the above, we can find:
\[ a = \sqrt{25} = 5, \quad b = \sqrt{20} = 2\sqrt{5} \]

**Step 3: Calculate the eccentricity \(e\)**
The eccentricity \(e\) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Calculating \(b^2\) and \(a^2\):
\[ b^2 = 20, \quad a^2 = 25 \]
Substituting these values:
\[ e = \sqrt{1 - \frac{20}{25}} = \sqrt{1 - 0.8} = \sqrt{0.2} = \frac{\sqrt{5}}{5} \]

**Step 4: Calculate the distance \(c\) from the center to the foci**
The distance \(c\) from the center to each focus is given by:
\[ c = ae \]
Substituting the value of \(a\):
\[ c = 5 \cdot \frac{\sqrt{5}}{5} = \sqrt{5} \]

**Step 5: Find the extremity of the minor axis**
The extremities of the minor axis are at \((0, \pm b)\), which gives us:
\[ (0, \pm 2\sqrt{5}) \]

**Step 6: Calculate the distance from the focus to the extremity of the minor axis**
The distance from one focus \((c, 0)\) to the extremity of the minor axis \((0, b)\) is:
\[ \text{Distance} = \sqrt{(c - 0)^2 + (0 - b)^2} = \sqrt{(\sqrt{5})^2 + (2\sqrt{5})^2} = \sqrt{5 + 20} = \sqrt{25} = 5 \]

### Final Answer for Part (i):
The distance between a focus and an extremity of the minor axis of the ellipse \(4x^2 + 5y^2 = 100\) is \(5\).

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### Part (ii): For the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

**Step 1: Identify values of \(a\) and \(b\)**
In this case, we have:
\[ a = a, \quad b = b \]

**Step 2: Calculate the eccentricity \(e\)**
The eccentricity \(e\) is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

**Step 3: Calculate the distance \(c\) from the center to the foci**
The distance \(c\) from the center to each focus is:
\[ c = ae \]

**Step 4: Find the extremity of the minor axis**
The extremities of the minor axis are at \((0, \pm b)\).

**Step 5: Calculate the distance from the focus to the extremity of the minor axis**
The distance from one focus \((c, 0)\) to the extremity of the minor axis \((0, b)\) is:
\[ \text{Distance} = \sqrt{(c - 0)^2 + (0 - b)^2} = \sqrt{(ae)^2 + b^2} \]
Substituting \(e\):
\[ = \sqrt{(a\sqrt{1 - \frac{b^2}{a^2}})^2 + b^2} = \sqrt{a^2(1 - \frac{b^2}{a^2}) + b^2} \]
This simplifies to:
\[ = \sqrt{a^2 - b^2 + b^2} = \sqrt{a^2} = a \]

### Final Answer for Part (ii):
The distance between a focus and an extremity of the minor axis of the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is \(a\).

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