Find the co-ordinates of that point on the curve `x^(2)/a^(2)+y^(2)/b^(2) = 1` at which the tangent drawn is parallel to Y-axis.

Answer

To find the coordinates of the point on the curve \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) where the tangent is parallel to the Y-axis, we can follow these steps:

### Step 1: Understand the condition for the tangent to be parallel to the Y-axis
A tangent line is parallel to the Y-axis when its slope is undefined, which occurs when the slope of the tangent is vertical. This means that the derivative \( \frac{dy}{dx} \) must be zero.

### Step 2: Differentiate the equation of the curve
We start with the equation of the curve:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}\left(\frac{x^2}{a^2}\right) + \frac{d}{dx}\left(\frac{y^2}{b^2}\right) = 0 \]
Using the chain rule, we get:
\[ \frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \]

### Step 3: Solve for \( \frac{dy}{dx} \)
Rearranging the equation, we find:
\[ \frac{2y}{b^2} \frac{dy}{dx} = -\frac{2x}{a^2} \]
Thus,
\[ \frac{dy}{dx} = -\frac{b^2 x}{a^2 y} \]

### Step 4: Set the derivative equal to zero
For the tangent to be vertical, we need \( \frac{dy}{dx} \) to be undefined, which occurs when \( y = 0 \) (since the denominator cannot be zero). Therefore, we set:
\[ y = 0 \]

### Step 5: Substitute \( y = 0 \) back into the curve equation
Now, substituting \( y = 0 \) into the original curve equation:
\[ \frac{x^2}{a^2} + \frac{0^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{x^2}{a^2} = 1 \]

### Step 6: Solve for \( x \)
Multiplying both sides by \( a^2 \):
\[ x^2 = a^2 \]
Taking the square root gives:
\[ x = a \quad \text{or} \quad x = -a \]

### Step 7: Write the coordinates
Thus, the coordinates of the points on the curve where the tangent is parallel to the Y-axis are:
\[ (a, 0) \quad \text{and} \quad (-a, 0) \]

### Final Answer
The coordinates of the points are \( (a, 0) \) and \( (-a, 0) \).

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