The co-ordinates of a focus of an ellipse is (4,0) and its eccentricity is `(4)/(5)` Its equation is :

Options

A. `25x^(2)+9y^(2)=225`
B. `9x^(2)+25y^(2)=225`
C. `16x^(2)+25y^(2)=400`
D. `25x^(2)+16y^(2)=400`.

Answer

To find the equation of the ellipse given the coordinates of a focus and its eccentricity, we can follow these steps:

### Step 1: Identify the given values
We have the focus of the ellipse at the point (4, 0) and the eccentricity \( e = \frac{4}{5} \).

### Step 2: Use the relationship between the focus and eccentricity
For an ellipse, the coordinates of the foci are given by \( (ae, 0) \) and \( (-ae, 0) \) when the ellipse is centered at the origin and aligned with the x-axis. Here, we have:
\[ ae = 4 \]

### Step 3: Solve for \( a \)
We know the eccentricity \( e = \frac{4}{5} \). We can substitute this into the equation \( ae = 4 \):
\[ a \cdot \frac{4}{5} = 4 \]
To find \( a \), multiply both sides by \( \frac{5}{4} \):
\[ a = 4 \cdot \frac{5}{4} = 5 \]

### Step 4: Use the eccentricity formula to find \( b \)
The relationship between \( a \), \( b \), and \( e \) is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting \( e = \frac{4}{5} \) and \( a = 5 \):
\[ \frac{4}{5} = \sqrt{1 - \frac{b^2}{25}} \]
Squaring both sides:
\[ \left(\frac{4}{5}\right)^2 = 1 - \frac{b^2}{25} \]
This simplifies to:
\[ \frac{16}{25} = 1 - \frac{b^2}{25} \]
Rearranging gives:
\[ \frac{b^2}{25} = 1 - \frac{16}{25} = \frac{9}{25} \]
Multiplying both sides by 25:
\[ b^2 = 9 \]
Taking the square root gives:
\[ b = 3 \]

### Step 5: Write the equation of the ellipse
The standard form of the equation of an ellipse centered at the origin with the major axis along the x-axis is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting \( a = 5 \) and \( b = 3 \):
\[ \frac{x^2}{5^2} + \frac{y^2}{3^2} = 1 \]
This simplifies to:
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]

### Final Answer
The equation of the ellipse is:
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]

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