What is the curve which passes through the point (1,1) and whose slope is `(2y)/(x)`?

Options

A. Circle
B. Parabola
C. Ellipse
D. Hyperbola

Answer

To solve the problem, we need to find the curve that passes through the point (1,1) and has a slope given by the expression \(\frac{2y}{x}\). We will follow these steps:

### Step 1: Set up the differential equation
We know that the slope of the curve at any point can be represented as the derivative \(\frac{dy}{dx}\). Given that the slope is \(\frac{2y}{x}\), we can write:
\[ \frac{dy}{dx} = \frac{2y}{x} \]

### Step 2: Rearrange the equation
To solve this differential equation, we can rearrange it:
\[ \frac{1}{y} \, dy = \frac{2}{x} \, dx \]

### Step 3: Integrate both sides
Now we will integrate both sides:
\[ \int \frac{1}{y} \, dy = \int \frac{2}{x} \, dx \]
The left side integrates to \(\ln |y|\) and the right side integrates to \(2 \ln |x| + C\):
\[ \ln |y| = 2 \ln |x| + C \]

### Step 4: Simplify the equation
Using properties of logarithms, we can rewrite the equation:
\[ \ln |y| = \ln |x|^2 + C \]
Exponentiating both sides gives:
\[ |y| = e^C |x|^2 \]
Let \(k = e^C\), then we have:
\[ y = kx^2 \]

### Step 5: Use the initial condition
We know that the curve passes through the point (1,1). We can use this to find the value of \(k\):
\[ 1 = k(1)^2 \implies k = 1 \]
Thus, the equation of the curve becomes:
\[ y = x^2 \]

### Conclusion
The curve that passes through the point (1,1) and has the slope \(\frac{2y}{x}\) is:
\[ y = x^2 \]
This is the equation of a parabola.