Find the co-ordinates of that point on the curve `x^(3)+y^(3)= a^(3)` at which the tangent drawn is parallel to X-axis.

Answer

To find the coordinates of the point on the curve \( x^3 + y^3 = a^3 \) at which the tangent drawn is parallel to the X-axis, we will follow these steps:

### Step 1: Understand the condition for the tangent to be parallel to the X-axis
A tangent line is parallel to the X-axis when its slope is zero. This means we need to find where the derivative \( \frac{dy}{dx} = 0 \).

### Step 2: Differentiate the given equation
We start with the equation of the curve:
\[ x^3 + y^3 = a^3 \]
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(a^3) \]
This gives us:
\[ 3x^2 + 3y^2 \frac{dy}{dx} = 0 \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Rearranging the equation, we isolate \(\frac{dy}{dx}\):
\[ 3y^2 \frac{dy}{dx} = -3x^2 \]
Dividing both sides by \(3y^2\) (assuming \(y \neq 0\)):
\[ \frac{dy}{dx} = -\frac{x^2}{y^2} \]

### Step 4: Set the derivative equal to zero
For the tangent to be parallel to the X-axis, we set the derivative equal to zero:
\[ -\frac{x^2}{y^2} = 0 \]
This implies that \(x^2 = 0\), which gives us:
\[ x = 0 \]

### Step 5: Substitute \(x = 0\) back into the original equation
Now we substitute \(x = 0\) back into the original curve equation to find \(y\):
\[ 0^3 + y^3 = a^3 \]
This simplifies to:
\[ y^3 = a^3 \]
Taking the cube root of both sides, we find:
\[ y = a \]

### Step 6: Write the coordinates of the point
Thus, the coordinates of the point on the curve where the tangent is parallel to the X-axis are:
\[ (0, a) \]

### Summary of the solution
The coordinates of the point on the curve \( x^3 + y^3 = a^3 \) at which the tangent is parallel to the X-axis are \( (0, a) \).

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