The number of normals to the curve `3y ^(3) =4x` which passes through the point `(0,1)` is

Answer

To find the number of normals to the curve \(3y^3 = 4x\) that pass through the point \((0, 1)\), we will follow these steps:

### Step 1: Differentiate the curve
We start with the equation of the curve:
\[ 3y^3 = 4x \]
To find the slope of the tangent line, we need to differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(3y^3) = \frac{d}{dx}(4x) \]
Using the chain rule on the left side, we get:
\[ 9y^2 \frac{dy}{dx} = 4 \]
Thus, the derivative (slope of the tangent) is:
\[ \frac{dy}{dx} = \frac{4}{9y^2} \]

### Step 2: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore:
\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = -\frac{9y^2}{4} \]

### Step 3: Equation of the normal line
The equation of the normal line at a point \((x_0, y_0)\) on the curve can be expressed using the point-slope form:
\[ y - y_0 = m(x - x_0) \]
Substituting the slope of the normal:
\[ y - y_0 = -\frac{9y_0^2}{4}(x - x_0) \]

### Step 4: Find the points on the curve
From the equation of the curve, we can express \(x\) in terms of \(y\):
\[ x = \frac{3y^3}{4} \]
Substituting \(x_0\) into the normal line equation, we get:
\[ y - y_0 = -\frac{9y_0^2}{4}\left(x - \frac{3y_0^3}{4}\right) \]

### Step 5: Substitute the point (0, 1)
Since the normal passes through the point \((0, 1)\), we substitute \(x = 0\) and \(y = 1\):
\[ 1 - y_0 = -\frac{9y_0^2}{4}\left(0 - \frac{3y_0^3}{4}\right) \]
This simplifies to:
\[ 1 - y_0 = \frac{27y_0^5}{16} \]
Rearranging gives:
\[ 27y_0^5 + 16y_0 - 16 = 0 \]

### Step 6: Solve the polynomial equation
We need to find the number of real roots of the polynomial:
\[ 27y_0^5 + 16y_0 - 16 = 0 \]
To determine the number of roots, we can use Descartes' Rule of Signs or analyze the function graphically or numerically.

### Step 7: Conclusion
After analyzing the polynomial, we find that there are 2 real roots. Therefore, the number of normals to the curve \(3y^3 = 4x\) that pass through the point \((0, 1)\) is **2**.

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