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.

Options

A. `(i) x=pm 6`
`(ii)y = pm 3 `
B. `(i) x=pm 2`
`(ii)y = pm 3 `
C. `(i) x=pm 4`
`(ii)y = pm 9 `
D. `(i) x=pm 4`
`(ii)y = pm 3 `

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 follow these steps:

### Step 1: Rewrite the equation of the ellipse
The given equation is:
\[ 9x^2 + 16y^2 = 144 \]
We can rewrite it in standard form by dividing through by 144:
\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]
This represents an ellipse centered at the origin with semi-major axis \(4\) (along the x-axis) and semi-minor axis \(3\) (along the y-axis).

### Step 2: Find the slope of the tangent
To find the slope of the tangent line, we differentiate the equation of the ellipse implicitly:
\[ \frac{d}{dx}(9x^2) + \frac{d}{dx}(16y^2) = \frac{d}{dx}(144) \]
This gives:
\[ 18x + 32y \frac{dy}{dx} = 0 \]
Rearranging for \(\frac{dy}{dx}\):
\[ 32y \frac{dy}{dx} = -18x \]
\[ \frac{dy}{dx} = -\frac{18x}{32y} = -\frac{9x}{16y} \]

### Step 3: Tangents parallel to the x-axis
For tangents parallel to the x-axis, the slope \(\frac{dy}{dx} = 0\). Thus, we set:
\[ -\frac{9x}{16y} = 0 \]
This implies \(x = 0\). Now we substitute \(x = 0\) back into the ellipse equation to find \(y\):
\[ 9(0)^2 + 16y^2 = 144 \implies 16y^2 = 144 \implies y^2 = 9 \implies y = \pm 3 \]
So, the points are \((0, 3)\) and \((0, -3)\).

The equations of the tangents at these points (where \(y = 3\) and \(y = -3\)) are:
\[ y = 3 \quad \text{and} \quad y = -3 \]

### Step 4: Tangents parallel to the y-axis
For tangents parallel to the y-axis, the slope is undefined, which occurs when the denominator in \(\frac{dy}{dx}\) is zero:
\[ 16y = 0 \implies y = 0 \]
Now we substitute \(y = 0\) back into the ellipse equation to find \(x\):
\[ 9x^2 + 16(0)^2 = 144 \implies 9x^2 = 144 \implies x^2 = 16 \implies x = \pm 4 \]
So, the points are \((4, 0)\) and \((-4, 0)\).

The equations of the tangents at these points (where \(x = 4\) and \(x = -4\)) are:
\[ x = 4 \quad \text{and} \quad x = -4 \]

### Final Result
The equations of the tangents to the curve at the specified points 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 \(x = -4\)

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