The points on the curve `(x^(2))/(9)+(y^(2))/(16)=1` at which the tangents are parallel to `y`- axis are:

Options

A. `(0, pm 4)`
B. `(pm 4, 0)`
C. `(pm 3, 0)`
D. `(0, pm 3)`

Answer

To find the points on the curve \(\frac{x^2}{9} + \frac{y^2}{16} = 1\) where the tangents are parallel to the y-axis, we can follow these steps:

### Step 1: Differentiate the equation of the curve
We start with the equation of the curve:
\[ \frac{x^2}{9} + \frac{y^2}{16} = 1 \]
Differentiating both sides with respect to \(x\) using implicit differentiation gives:
\[ \frac{d}{dx}\left(\frac{x^2}{9}\right) + \frac{d}{dx}\left(\frac{y^2}{16}\right) = 0 \]
This results in:
\[ \frac{2x}{9} + \frac{2y}{16} \cdot \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ \frac{2y}{16} \cdot \frac{dy}{dx} = -\frac{2x}{9} \]
Dividing both sides by \(\frac{2y}{16}\) gives:
\[ \frac{dy}{dx} = -\frac{16x}{9y} \]

### Step 3: Determine when the tangent is parallel to the y-axis
A tangent line is parallel to the y-axis when its slope is undefined, which occurs when the denominator of \(\frac{dy}{dx}\) is zero. Thus, we set:
\[ y = 0 \]

### Step 4: Substitute \(y = 0\) back into the curve equation
Now we substitute \(y = 0\) back into the original equation:
\[ \frac{x^2}{9} + \frac{0^2}{16} = 1 \]
This simplifies to:
\[ \frac{x^2}{9} = 1 \]

### Step 5: Solve for \(x\)
Multiplying both sides by 9 gives:
\[ x^2 = 9 \]
Taking the square root of both sides results in:
\[ x = \pm 3 \]

### Step 6: Write the coordinates of the points
The points on the curve where the tangents are parallel to the y-axis are:
\[ (3, 0) \quad \text{and} \quad (-3, 0) \]

### Final Answer
Thus, the points on the curve where the tangents are parallel to the y-axis are:
\[ (3, 0) \quad \text{and} \quad (-3, 0) \]

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