A family of curves is such that the length intercepted on the y-axis between the origin and the tangent at a point is three times the ordinate of the point of contact. The family of curves is

Options

A. xy, =c,c is a constant
B. `xy^(2)` =c,c is a constant
C. `x^(2)y ` = c, c is a constant
D. `x^(2)y^(2)` = c,c is a constant

Answer

To solve the problem, we need to find the family of curves based on the given condition. Let's break down the solution step by step.

### Step 1: Understand the given condition
We are given that the length intercepted on the y-axis between the origin and the tangent at a point is three times the ordinate (y-coordinate) of the point of contact.

### Step 2: Set up the point on the curve
Let \( (h, k) \) be a point on the curve, where \( k \) is the ordinate (y-coordinate) and \( h \) is the abscissa (x-coordinate).

### Step 3: Find the slope of the tangent
Let \( \frac{dy}{dx} = m \) be the slope of the tangent at the point \( (h, k) \).

### Step 4: Write the equation of the tangent
The equation of the tangent line at the point \( (h, k) \) can be expressed as:
\[ y - k = m(x - h) \]

### Step 5: Find the y-intercept of the tangent
To find the y-intercept, set \( x = 0 \):
\[ y - k = m(0 - h) \implies y = k - mh \]
Thus, the y-intercept is \( k - mh \).

### Step 6: Relate the y-intercept to the ordinate
According to the problem, the length intercepted on the y-axis (which is \( k - mh \)) is three times the ordinate \( k \):
\[ k - mh = 3k \]
This simplifies to:
\[ -kh = 2k \implies -mh = 2k \]

### Step 7: Express \( m \) in terms of \( k \) and \( h \)
From the equation \( -mh = 2k \), we can express \( m \):
\[ m = -\frac{2k}{h} \]

### Step 8: Replace \( h \) and \( k \) with \( x \) and \( y \)
Since \( (h, k) \) is any point on the curve, we can replace \( h \) with \( x \) and \( k \) with \( y \):
\[ \frac{dy}{dx} = -\frac{2y}{x} \]

### Step 9: Separate variables and integrate
We can separate the variables:
\[ \frac{dy}{y} = -2 \frac{dx}{x} \]
Now, integrate both sides:
\[ \int \frac{dy}{y} = -2 \int \frac{dx}{x} \]
This gives:
\[ \ln |y| = -2 \ln |x| + C \]

### Step 10: Exponentiate to eliminate the logarithm
Exponentiating both sides, we have:
\[ |y| = e^{C} |x|^{-2} \]
Let \( e^{C} = k \) (where \( k \) is a constant), so:
\[ y = \frac{k}{x^2} \]

### Step 11: Rearranging the equation
We can rearrange this to:
\[ x^2y = k \]
Thus, the family of curves is given by:
\[ x^2y = C' \]
where \( C' \) is a constant.

### Conclusion
The family of curves is \( x^2y = C' \).

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