If the straight line `4ax+3by=24` is a normal to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 (agtb)`, then find the the coordinates of focii and the ellipse

Explanation

To solve the problem, we need to find the coordinates of the foci of the ellipse given that the line \(4ax + 3by = 24\) is a normal to the ellipse defined by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) where \(a > b\).

### Step 1: Write down the given normal equation
The equation of the normal line is given as:
\[ 4ax + 3by = 24 \]
Let's label this as Equation (1).

### Step 2: Identify the general point on the ellipse
A general point on the ellipse can be represented as:
\[ P(a \cos \theta, b \sin \theta) \]
where \(\theta\) is a parameter.

### Step 3: Write down the general equation of the normal to the ellipse
The general equation of the normal to the ellipse at point \(P(x_1, y_1)\) is given by:
\[ \frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2 \]
Substituting \(x_1 = a \cos \theta\) and \(y_1 = b \sin \theta\), we have:
\[ \frac{a^2 x}{a \cos \theta} - \frac{b^2 y}{b \sin \theta} = a^2 - b^2 \]
This simplifies to:
\[ \frac{a x}{\cos \theta} - \frac{b y}{\sin \theta} = a^2 - b^2 \]
Let's label this as Equation (2).

### Step 4: Equate the two equations
Since the line \(4ax + 3by = 24\) is a normal to the ellipse, Equations (1) and (2) must be identical. Therefore, we can equate the coefficients of \(x\), \(y\), and the constant terms.

From Equation (1):
- Coefficient of \(x\) is \(4a\)
- Coefficient of \(y\) is \(3b\)
- Constant term is \(24\)

From Equation (2):
- Coefficient of \(x\) is \(\frac{a}{\cos \theta}\)
- Coefficient of \(y\) is \(-\frac{b}{\sin \theta}\)
- Constant term is \(a^2 - b^2\)

### Step 5: Set up the equations
From the coefficients, we have:
1. \(\frac{a}{\cos \theta} = 4a\)
2. \(-\frac{b}{\sin \theta} = 3b\)
3. \(a^2 - b^2 = 24\)

### Step 6: Solve for \(\theta\)
From the first equation:
\[ \cos \theta = \frac{1}{4} \]
From the second equation:
\[ \sin \theta = -\frac{1}{3} \]

### Step 7: Use Pythagorean identity
We know that:
\[ \cos^2 \theta + \sin^2 \theta = 1 \]
Substituting the values:
\[ \left(\frac{1}{4}\right)^2 + \left(-\frac{1}{3}\right)^2 = 1 \]
Calculating:
\[ \frac{1}{16} + \frac{1}{9} = \frac{9}{144} + \frac{16}{144} = \frac{25}{144} \neq 1 \]
This indicates that we need to check our values for \(\theta\).

### Step 8: Find \(a^2 - b^2\)
From the third equation:
\[ a^2 - b^2 = 24 \]

### Step 9: Use the relationship between \(a\) and \(b\)
Since \(a > b\), we can express \(b^2\) in terms of \(a^2\):
\[ b^2 = a^2 - 24 \]

### Step 10: Find the coordinates of the foci
The foci of the ellipse are given by the formula:
\[ (\pm ae, 0) \]
where \(e = \sqrt{1 - \frac{b^2}{a^2}}\).

Substituting \(b^2 = a^2 - 24\):
\[ e = \sqrt{1 - \frac{a^2 - 24}{a^2}} = \sqrt{\frac{24}{a^2}} \]
Thus, the coordinates of the foci are:
\[ \left(\pm a \sqrt{\frac{24}{a^2}}, 0\right) = \left(\pm \sqrt{24}, 0\right) = \left(\pm 2\sqrt{6}, 0\right) \]

### Final Answer
The coordinates of the foci of the ellipse are:
\[ (2\sqrt{6}, 0) \text{ and } (-2\sqrt{6}, 0) \]