The equation of the curve which passes through the point (2a, a) and for which the sum of the cartesian sub tangent and the abscissa is equal to the constant a , is

Options

A. `y(x-a)=a^(2) `
B. `y(x+a)=a^(2) `
C. `x(y-a)=a^(2) `
D. `x(y+a)=a^(2)`

Answer

To find the equation of the curve that passes through the point \((2a, a)\) and satisfies the condition that the sum of the Cartesian sub-tangent and the abscissa is equal to the constant \(a\), we can follow these steps:

### Step 1: Understand the Condition
The Cartesian sub-tangent \(T\) at any point on the curve is given by the formula:
\[ T = \frac{y}{\frac{dy}{dx}} \]
According to the problem, we have:
\[ T + x = a \]
Substituting the expression for \(T\), we get:
\[ \frac{y}{\frac{dy}{dx}} + x = a \]

### Step 2: Rearranging the Equation
Rearranging the equation gives us:
\[ \frac{y}{\frac{dy}{dx}} = a - x \]
This can be rewritten as:
\[ \frac{dy}{dx} = \frac{y}{a - x} \]

### Step 3: Separating Variables
Now, we can separate the variables:
\[ \frac{dy}{y} = \frac{dx}{a - x} \]

### Step 4: Integrating Both Sides
Integrating both sides, we have:
\[ \int \frac{dy}{y} = \int \frac{dx}{a - x} \]
This gives us:
\[ \ln |y| = -\ln |a - x| + C \]
where \(C\) is the constant of integration.

### Step 5: Simplifying the Equation
We can exponentiate both sides to eliminate the logarithm:
\[ |y| = e^{C} \cdot \frac{1}{|a - x|} \]
Let \(k = e^{C}\), then:
\[ y = \frac{k}{a - x} \]

### Step 6: Applying the Initial Condition
Now we apply the initial condition that the curve passes through the point \((2a, a)\):
\[ a = \frac{k}{a - 2a} \]
This simplifies to:
\[ a = \frac{k}{-a} \implies k = -a^2 \]

### Step 7: Final Equation of the Curve
Substituting \(k\) back into the equation for \(y\):
\[ y = \frac{-a^2}{a - x} \]
This can be rearranged to:
\[ y(a - x) = -a^2 \]
or:
\[ y(x - a) = a^2 \]

Thus, the required equation of the curve is:
\[ y(x - a) = a^2 \]