find the length of major and minor of exis of the following ellipse, `16x^2+25y^2=400`

Answer

To find the lengths of the major and minor axes of the ellipse given by the equation \(16x^2 + 25y^2 = 400\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
The standard form of an ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
To convert the given equation \(16x^2 + 25y^2 = 400\) into this form, we divide every term by 400.

\[ \frac{16x^2}{400} + \frac{25y^2}{400} = 1 \]

### Step 2: Simplify the fractions
Now, simplify the fractions:
\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]

### Step 3: Identify \(a^2\) and \(b^2\)
From the equation \(\frac{x^2}{25} + \frac{y^2}{16} = 1\), we can identify:
- \(a^2 = 25\)
- \(b^2 = 16\)

### Step 4: Calculate \(a\) and \(b\)
Now, we take the square roots to find \(a\) and \(b\):
\[ a = \sqrt{25} = 5 \]
\[ b = \sqrt{16} = 4 \]

### Step 5: Calculate the lengths of the axes
The lengths of the major and minor axes are given by:
- Length of the major axis = \(2a\)
- Length of the minor axis = \(2b\)

Calculating these:
\[ \text{Length of the major axis} = 2a = 2 \times 5 = 10 \]
\[ \text{Length of the minor axis} = 2b = 2 \times 4 = 8 \]

### Final Answer
- Length of the major axis = 10 units
- Length of the minor axis = 8 units

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