Find the equation of tangent of the curve `9x^(2)+16y^(2) = 144` at those points at which tangents are parallel to (i) X-axis, (ii) Y-axis.

Answer

To find the equations of the tangents to the curve \(9x^2 + 16y^2 = 144\) at points where the tangents are parallel to the x-axis and y-axis, we will follow these steps:

### Step 1: Differentiate the given equation
The given equation is:
\[ 9x^2 + 16y^2 = 144 \]
We will differentiate this implicitly with respect to \(x\).

Differentiating both sides:
\[ \frac{d}{dx}(9x^2) + \frac{d}{dx}(16y^2) = \frac{d}{dx}(144) \]
This gives:
\[ 18x + 32y \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ 32y \frac{dy}{dx} = -18x \]
\[ \frac{dy}{dx} = -\frac{18x}{32y} = -\frac{9x}{16y} \]

### Step 3: Find points where tangents are parallel to the x-axis
For the tangent to be parallel to the x-axis, the slope \(\frac{dy}{dx}\) must be \(0\):
\[ -\frac{9x}{16y} = 0 \]
This implies:
\[ 9x = 0 \quad \Rightarrow \quad x = 0 \]

### Step 4: Substitute \(x = 0\) into the original equation
Substituting \(x = 0\) into the original equation:
\[ 9(0)^2 + 16y^2 = 144 \]
This simplifies to:
\[ 16y^2 = 144 \quad \Rightarrow \quad y^2 = 9 \quad \Rightarrow \quad y = \pm 3 \]
Thus, the points are \((0, 3)\) and \((0, -3)\).

### Step 5: Write the equations of the tangents at these points
The tangents at these points are horizontal lines (since they are parallel to the x-axis):
- At \((0, 3)\): \(y = 3\)
- At \((0, -3)\): \(y = -3\)

So, the equations of the tangents parallel to the x-axis are:
\[ y = 3 \quad \text{and} \quad y = -3 \]

### Step 6: Find points where tangents are parallel to the y-axis
For the tangent to be parallel to the y-axis, the slope \(\frac{dy}{dx}\) must be undefined (infinity), which occurs when the denominator \(16y\) is \(0\):
\[ 16y = 0 \quad \Rightarrow \quad y = 0 \]

### Step 7: Substitute \(y = 0\) into the original equation
Substituting \(y = 0\) into the original equation:
\[ 9x^2 + 16(0)^2 = 144 \]
This simplifies to:
\[ 9x^2 = 144 \quad \Rightarrow \quad x^2 = 16 \quad \Rightarrow \quad x = \pm 4 \]
Thus, the points are \((4, 0)\) and \((-4, 0)\).

### Step 8: Write the equations of the tangents at these points
The tangents at these points are vertical lines (since they are parallel to the y-axis):
- At \((4, 0)\): \(x = 4\)
- At \((-4, 0)\): \(x = -4\)

So, the equations of the tangents parallel to the y-axis are:
\[ x = 4 \quad \text{and} \quad x = -4 \]

### Final Answer
The equations of the tangents are:
1. For tangents parallel to the x-axis: \(y = 3\) and \(y = -3\)
2. For tangents parallel to the y-axis: \(x = 4\) and \(x = -4\)