The equation of the ellipse whose centre is at origin, major axis is along x-axis with eccentricity `(3)/(4)` and latus rectum 4 units is:

Options

A. A. `(x^(2))/(1024)+(7y^(2))/(64)=1`
B. B. `(49x^(2))/(1024)+(7y^(2))/(64)=1`
C. C. `(7x^(2))/(1024)+(49y^(2))/(64)=1`
D. D. `(x^(2))/(1024)+(y^(2))/(64)=1`

Answer

To find the equation of the ellipse with the given parameters, we can follow these steps:

### Step 1: Understand the standard form of the ellipse
The standard form of the equation of an ellipse with its center at the origin and the major axis along the x-axis is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

### Step 2: Use the eccentricity to find the relationship between \( a \) and \( b \)
The eccentricity \( e \) of the ellipse is related to \( a \) and \( b \) by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Squaring both sides gives:
\[ e^2 = 1 - \frac{b^2}{a^2} \]
Rearranging this, we find:
\[ \frac{b^2}{a^2} = 1 - e^2 \]
Thus,
\[ b^2 = a^2(1 - e^2) \]

### Step 3: Substitute the given eccentricity
Given that \( e = \frac{3}{4} \):
\[ e^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \]
Then,
\[ 1 - e^2 = 1 - \frac{9}{16} = \frac{7}{16} \]
Thus,
\[ b^2 = a^2 \cdot \frac{7}{16} \]

### Step 4: Use the latus rectum to find another relationship
The length of the latus rectum \( L \) of an ellipse is given by:
\[ L = \frac{2b^2}{a} \]
We are given \( L = 4 \):
\[ 4 = \frac{2b^2}{a} \]
Substituting \( b^2 = a^2 \cdot \frac{7}{16} \):
\[ 4 = \frac{2 \cdot \frac{7}{16} a^2}{a} \]
This simplifies to:
\[ 4 = \frac{14a}{16} = \frac{7a}{8} \]
Multiplying both sides by 8:
\[ 32 = 7a \]
Thus,
\[ a = \frac{32}{7} \]

### Step 5: Find \( b^2 \)
Now substituting \( a \) back to find \( b^2 \):
\[ b^2 = a^2 \cdot \frac{7}{16} = \left(\frac{32}{7}\right)^2 \cdot \frac{7}{16} \]
Calculating \( a^2 \):
\[ a^2 = \frac{1024}{49} \]
Now substituting:
\[ b^2 = \frac{1024}{49} \cdot \frac{7}{16} = \frac{1024 \cdot 7}{49 \cdot 16} \]
Calculating \( 49 \cdot 16 = 784 \):
\[ b^2 = \frac{7168}{784} = \frac{896}{98} = \frac{448}{49} \]

### Step 6: Write the equation of the ellipse
Now substituting \( a^2 \) and \( b^2 \) into the standard form:
\[ \frac{x^2}{\frac{1024}{49}} + \frac{y^2}{\frac{448}{49}} = 1 \]
This simplifies to:
\[ \frac{49x^2}{1024} + \frac{49y^2}{448} = 1 \]
Multiplying through by 1024 gives:
\[ 49x^2 + \frac{49 \cdot 1024}{448}y^2 = 1024 \]
Thus, the final equation is:
\[ \frac{x^2}{\frac{1024}{49}} + \frac{y^2}{\frac{448}{49}} = 1 \]

### Final Answer
The equation of the ellipse is:
\[ \frac{x^2}{1024} + \frac{y^2}{448} = 1 \]