A curve passes through the point (5, 3) and at any point (x,y) on it, the product of its slope and the ordinate is equal to its abscissa. Its equation is given by

Options

A. `x^(2)+y^(2)=34`
B. `x^(2)-y^(2)=16`
C. `5y-3x=0`
D. None of these

Answer

To solve the problem step by step, we will derive the equation of the curve based on the given conditions.

### Step 1: Understand the condition
We are given that at any point \((x, y)\) on the curve, the product of its slope (which is \(\frac{dy}{dx}\)) and the ordinate (which is \(y\)) is equal to its abscissa (which is \(x\)). This can be expressed mathematically as:
\[ y \frac{dy}{dx} = x \]

### Step 2: Rearranging the equation
We can rearrange the equation to separate the variables:
\[ y \, dy = x \, dx \]

### Step 3: Integrate both sides
Now, we will integrate both sides:
\[ \int y \, dy = \int x \, dx \]
This gives us:
\[ \frac{y^2}{2} = \frac{x^2}{2} + C \]
where \(C\) is the constant of integration.

### Step 4: Multiply through by 2
To simplify, we can multiply the entire equation by 2:
\[ y^2 = x^2 + 2C \]

### Step 5: Use the point (5, 3) to find \(C\)
We know that the curve passes through the point \((5, 3)\). We can substitute \(x = 5\) and \(y = 3\) into the equation to find \(C\):
\[ 3^2 = 5^2 + 2C \]
This simplifies to:
\[ 9 = 25 + 2C \]
Now, solving for \(2C\):
\[ 2C = 9 - 25 = -16 \]

### Step 6: Substitute back to find the equation
Now we substitute \(2C = -16\) back into the equation:
\[ y^2 = x^2 - 16 \]

### Final Equation
Thus, the equation of the curve is:
\[ y^2 = x^2 - 16 \]