Tangents to the folium of descartes `x^3 + y^3` = 3axy at the point where it meets the parabola `y^2 = ax` are parallel to

Options

A. x-axis
B. y-axis
C. y = x
D. none

Answer

To solve the problem of finding the tangents to the folium of Descartes given by the equation \( x^3 + y^3 = 3axy \) at the points where it meets the parabola \( y^2 = ax \), we will follow these steps:

### Step 1: Find the points of intersection
We start by substituting \( x \) from the parabola equation into the folium equation. The parabola is given by:
\[ y^2 = ax \implies x = \frac{y^2}{a} \]
Now, substituting this into the folium equation:
\[ \left(\frac{y^2}{a}\right)^3 + y^3 = 3a\left(\frac{y^2}{a}\right)y \]
This simplifies to:
\[ \frac{y^6}{a^3} + y^3 = 3y^3 \]
Rearranging gives:
\[ \frac{y^6}{a^3} - 2y^3 = 0 \]
Factoring out \( y^3 \):
\[ y^3\left(\frac{y^3}{a^3} - 2\right) = 0 \]
Thus, \( y^3 = 0 \) or \( \frac{y^3}{a^3} - 2 = 0 \). This gives us:
1. \( y = 0 \) (which gives \( x = 0 \))
2. \( y^3 = 2a^3 \implies y = \sqrt[3]{2a^3} = a\sqrt[3]{2} \) (which gives \( x = \frac{(a\sqrt[3]{2})^2}{a} = a^{1/3} \cdot 2^{2/3} \))

### Step 2: Find the slope of the tangent
Next, we need to find the slope of the tangent to the folium at the points of intersection. We differentiate the folium equation implicitly:
\[ 3x^2 + 3y^2 \frac{dy}{dx} = 3ay + 3x \frac{dy}{dx} \]
Rearranging gives:
\[ 3y^2 \frac{dy}{dx} - 3x \frac{dy}{dx} = 3ay - 3x^2 \]
Factoring out \( \frac{dy}{dx} \):
\[ \frac{dy}{dx}(3y^2 - 3x) = 3ay - 3x^2 \]
Thus:
\[ \frac{dy}{dx} = \frac{3ay - 3x^2}{3y^2 - 3x} = \frac{ay - x^2}{y^2 - x} \]

### Step 3: Evaluate the slope at the points of intersection
Now we evaluate \( \frac{dy}{dx} \) at the intersection points.

1. At the point \( (0, 0) \):
\[ \frac{dy}{dx} = \frac{a(0) - 0^2}{0^2 - 0} \implies \text{undefined (infinite slope)} \]

2. At the point \( \left(a^{1/3} \cdot 2^{2/3}, a \sqrt[3]{2}\right) \):
Substitute \( x = a^{1/3} \cdot 2^{2/3} \) and \( y = a \sqrt[3]{2} \):
\[ \frac{dy}{dx} = \frac{a(a\sqrt[3]{2}) - (a^{1/3} \cdot 2^{2/3})^2}{(a\sqrt[3]{2})^2 - (a^{1/3} \cdot 2^{2/3})} \]
This will yield a finite slope.

### Step 4: Determine the nature of the tangents
Since the slope at the point \( (0, 0) \) is infinite, the tangent line is vertical, which means it is parallel to the y-axis.

### Conclusion
The tangents to the folium of Descartes at the points where it meets the parabola are parallel to the y-axis.