The point on the curve `3y = 6x-5x^3` the normal at Which passes through the origin, is

Options

A. `(1,1//3)`
B. `(-1,-1//3)`
C. `(2,-28//3)`
D. none of these

Answer

To find the point on the curve \(3y = 6x - 5x^3\) where the normal at that point passes through the origin, we can follow these steps:

### Step 1: Rewrite the equation of the curve
The given equation of the curve is:
\[ 3y = 6x - 5x^3 \]
We can express \(y\) in terms of \(x\):
\[ y = \frac{6x - 5x^3}{3} \]

### Step 2: Differentiate the curve
Next, we differentiate \(y\) with respect to \(x\) to find the slope of the tangent line:
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{6x - 5x^3}{3}\right) = \frac{1}{3}(6 - 15x^2) = 2 - 5x^2 \]

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{2 - 5x^2} \]

### Step 4: Write the equation of the normal
Let \((x_1, y_1)\) be the point on the curve where the normal passes through the origin. The equation of the normal line at this point can be written as:
\[ y - y_1 = -\frac{1}{2 - 5x_1^2}(x - x_1) \]

### Step 5: Substitute the origin into the normal equation
Since the normal passes through the origin \((0, 0)\), we can substitute \(x = 0\) and \(y = 0\) into the normal equation:
\[ 0 - y_1 = -\frac{1}{2 - 5x_1^2}(0 - x_1) \]
This simplifies to:
\[ -y_1 = \frac{x_1}{2 - 5x_1^2} \]
Thus, we have:
\[ y_1 = -\frac{x_1}{2 - 5x_1^2} \]

### Step 6: Substitute \(y_1\) back into the curve equation
We know that \(y_1\) must also satisfy the original curve equation:
\[ 3y_1 = 6x_1 - 5x_1^3 \]
Substituting for \(y_1\):
\[ 3\left(-\frac{x_1}{2 - 5x_1^2}\right) = 6x_1 - 5x_1^3 \]
This simplifies to:
\[ -\frac{3x_1}{2 - 5x_1^2} = 6x_1 - 5x_1^3 \]

### Step 7: Clear the fraction
Multiply both sides by \(2 - 5x_1^2\) to eliminate the denominator:
\[ -3x_1 = (6x_1 - 5x_1^3)(2 - 5x_1^2) \]
Expanding the right-hand side:
\[ -3x_1 = 12x_1 - 30x_1^3 - 10x_1^3 + 25x_1^5 \]
This simplifies to:
\[ -3x_1 = 12x_1 - 40x_1^3 + 25x_1^5 \]
Rearranging gives:
\[ 25x_1^5 - 40x_1^3 + 15x_1 = 0 \]
Factoring out \(x_1\):
\[ x_1(25x_1^4 - 40x_1^2 + 15) = 0 \]

### Step 8: Solve for \(x_1\)
This gives us two cases:
1. \(x_1 = 0\) (which we will discard as it does not give a normal passing through the origin)
2. Solve \(25x_1^4 - 40x_1^2 + 15 = 0\) by substituting \(u = x_1^2\):
\[ 25u^2 - 40u + 15 = 0 \]
Using the quadratic formula:
\[ u = \frac{40 \pm \sqrt{(-40)^2 - 4 \cdot 25 \cdot 15}}{2 \cdot 25} \]
Calculating the discriminant:
\[ 1600 - 1500 = 100 \]
Thus:
\[ u = \frac{40 \pm 10}{50} \]
This gives:
\[ u = 1 \quad \text{or} \quad u = \frac{3}{5} \]
So:
\[ x_1^2 = 1 \implies x_1 = 1 \quad \text{or} \quad x_1^2 = \frac{3}{5} \implies x_1 = \sqrt{\frac{3}{5}} \]

### Step 9: Find corresponding \(y_1\)
For \(x_1 = 1\):
\[ y_1 = -\frac{1}{2 - 5 \cdot 1^2} = -\frac{1}{-3} = \frac{1}{3} \]

For \(x_1 = \sqrt{\frac{3}{5}}\):
\[ y_1 = -\frac{\sqrt{\frac{3}{5}}}{2 - 5 \cdot \frac{3}{5}} = -\frac{\sqrt{\frac{3}{5}}}{2 - 3} = \sqrt{\frac{3}{5}} \]

### Final Points
Thus, the points on the curve where the normal passes through the origin are:
1. \((1, \frac{1}{3})\)
2. \((\sqrt{\frac{3}{5}}, \sqrt{\frac{3}{5}})\)

However, since the problem specifically asks for the point on the curve where the normal passes through the origin, we conclude that the answer is:
\[ \text{The point is } (1, \frac{1}{3}). \]