The equation(s) of the tangent(s) to the curve `y=x^(4)` from the point (2, 0) not on the curve is given by

Options

A. y=0
B. y-1=5(x-1)
C. `y-(4096)/(81)=(2048)/(27)(x-(8)/(3))`
D. `y-(32)/(243)=(80)/(81)(x-(2)/(3))`

Answer

To find the equations of the tangents to the curve \( y = x^4 \) from the point \( (2, 0) \), we will follow these steps:

### Step 1: Find the derivative of the curve
The first step is to find the derivative of the curve \( y = x^4 \) to determine the slope of the tangent line at any point on the curve.

\[ \frac{dy}{dx} = 4x^3 \]

### Step 2: Set up the equation of the tangent line
The equation of the tangent line at a point \( (a, a^4) \) on the curve can be expressed using the point-slope form of a line:

\[ y - a^4 = m(x - a) \]

where \( m \) is the slope of the tangent line at point \( (a, a^4) \), which is \( m = 4a^3 \).

Substituting the slope into the equation gives:

\[ y - a^4 = 4a^3(x - a) \]

### Step 3: Substitute the point (2, 0) into the tangent equation
Since the tangent line must pass through the point \( (2, 0) \), we substitute \( x = 2 \) and \( y = 0 \) into the tangent equation:

\[ 0 - a^4 = 4a^3(2 - a) \]

This simplifies to:

\[ -a^4 = 4a^3(2 - a) \]

### Step 4: Rearranging the equation
Rearranging the equation gives:

\[ -a^4 = 8a^3 - 4a^4 \]

Combining like terms results in:

\[ 3a^4 - 8a^3 = 0 \]

### Step 5: Factor the equation
Factoring out \( a^3 \):

\[ a^3(3a - 8) = 0 \]

This gives us two possible solutions:

1. \( a^3 = 0 \) which implies \( a = 0 \)
2. \( 3a - 8 = 0 \) which implies \( a = \frac{8}{3} \)

### Step 6: Find the points on the curve
Now we find the corresponding points on the curve for these values of \( a \):

1. For \( a = 0 \):
\[ (0, 0^4) = (0, 0) \]

2. For \( a = \frac{8}{3} \):
\[ \left(\frac{8}{3}, \left(\frac{8}{3}\right)^4\right) = \left(\frac{8}{3}, \frac{4096}{81}\right) \]

### Step 7: Write the equations of the tangents
Now we can write the equations of the tangents for both points:

1. For the point \( (0, 0) \):
\[ y - 0 = 0(x - 0) \implies y = 0 \]

2. For the point \( \left(\frac{8}{3}, \frac{4096}{81}\right) \):
The slope at this point is:
\[ m = 4\left(\frac{8}{3}\right)^3 = \frac{2048}{27} \]
The tangent line equation becomes:
\[ y - \frac{4096}{81} = \frac{2048}{27}\left(x - \frac{8}{3}\right) \]

### Final Tangent Equations
Thus, the equations of the tangents to the curve \( y = x^4 \) from the point \( (2, 0) \) are:

1. \( y = 0 \)
2. \( y - \frac{4096}{81} = \frac{2048}{27}\left(x - \frac{8}{3}\right) \)