The point on the curve `6y=4x^(3)-3x^(2)`, the tangent at which makes an equal angle with the coordinate axes is

Options

A. `(1, -(1)/(6))`
B. `(-1, -(7)/(6))`
C. `(-(1)/(2), -(5)/(24))`
D. `((1)/(2), -(1)/(24))`

Answer

To find the point on the curve \(6y = 4x^3 - 3x^2\) where the tangent makes an equal angle with the coordinate axes, we can follow these steps:

### Step 1: Rewrite the equation of the curve
The given equation of the curve is:
\[ 6y = 4x^3 - 3x^2 \]
We can express \(y\) in terms of \(x\):
\[ y = \frac{4x^3 - 3x^2}{6} = \frac{2x^3 - \frac{1}{2}x^2}{3} \]

### Step 2: Find the derivative
To find the slope of the tangent, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{4x^3 - 3x^2}{6}\right) \]
Using the power rule:
\[ \frac{dy}{dx} = \frac{1}{6}(12x^2 - 6x) = \frac{2x^2 - x}{1} \]

### Step 3: Set the slope equal to ±1
Since the tangent makes an equal angle with the coordinate axes, the slope of the tangent line must be \(1\) or \(-1\). Therefore, we set:
\[ 2x^2 - x = 1 \quad \text{(1)} \]
and
\[ 2x^2 - x = -1 \quad \text{(2)} \]

### Step 4: Solve the equations
**For equation (1):**
\[ 2x^2 - x - 1 = 0 \]
Factoring:
\[ (2x + 1)(x - 1) = 0 \]
Thus, \(x = 1\) or \(x = -\frac{1}{2}\).

**For equation (2):**
\[ 2x^2 - x + 1 = 0 \]
Calculating the discriminant:
\[ b^2 - 4ac = (-1)^2 - 4(2)(1) = 1 - 8 = -7 \]
Since the discriminant is negative, this equation has no real solutions.

### Step 5: Find corresponding \(y\) values
Now we will find \(y\) for the valid \(x\) values from equation (1).

**For \(x = 1\):**
\[ 6y = 4(1)^3 - 3(1)^2 = 4 - 3 = 1 \implies y = \frac{1}{6} \]
So, the point is \((1, \frac{1}{6})\).

**For \(x = -\frac{1}{2}\):**
\[ 6y = 4\left(-\frac{1}{2}\right)^3 - 3\left(-\frac{1}{2}\right)^2 \]
Calculating:
\[ 6y = 4\left(-\frac{1}{8}\right) - 3\left(\frac{1}{4}\right) = -\frac{1}{2} - \frac{3}{4} = -\frac{2}{4} - \frac{3}{4} = -\frac{5}{4} \]
Thus,
\[ y = -\frac{5}{24} \]
So, the point is \(\left(-\frac{1}{2}, -\frac{5}{24}\right)\).

### Conclusion
The points on the curve where the tangent makes an equal angle with the coordinate axes are:
1. \((1, \frac{1}{6})\)
2. \(\left(-\frac{1}{2}, -\frac{5}{24}\right)\)