Question

A curve passes through the point $(3, 2)$ for which the segment of the tangent line contained between the coordinate axes is bisected at the point of contact. The equation of the curve is

• A. y=x2-7
• B. x=y2/2+2
• C. xy=6
• D. x2+y2-5x+7y+11=0

Answer 1

The Correct Option is C

Let the curve be \( y = f(x) \), and suppose it passes through the point \( (x, y) \), with the tangent at this point having slope \( m = \frac{dy}{dx} \).

The general equation of the tangent line at point \( (x, y) \) is: \[ Y - y = m(X - x) \]

Let’s find where this tangent cuts the x-axis and y-axis.

• x-intercept: set \( Y = 0 \) in the equation: \[ 0 - y = m(X - x) \Rightarrow X = x - \frac{y}{m} \]
• y-intercept: set \( X = 0 \): \[ Y - y = m(0 - x) \Rightarrow Y = y - mx \]

So the intercepts are: \[ A = \left(x - \frac{y}{m}, 0\right), \quad B = \left(0, y - mx\right) \]

Now consider the segment AB. The midpoint of this segment is: \[ \left( \frac{x - \frac{y}{m}}{2}, \frac{y - mx}{2} \right) \]

We are given that the point of contact \( (x, y) \) is the midpoint. So equating coordinates:

\[ x = \frac{x - \frac{y}{m}}{2}, \quad y = \frac{y - mx}{2} \]

Now solve both equations. Start with the first:

Multiply both sides by 2: \[ 2x = x - \frac{y}{m} \Rightarrow x = -\frac{y}{m} \Rightarrow m = -\frac{y}{x} \]

Now use this in the second equation:

\[ y = \frac{y - mx}{2} \Rightarrow 2y = y - mx \Rightarrow y = -mx \]

Substitute \( m = -\frac{y}{x} \): \[ y = -\left(-\frac{y}{x}\right)x = y \] So both conditions are satisfied.

So we conclude: \[ \frac{dy}{dx} = m = -\frac{y}{x} \Rightarrow \frac{dy}{dx} = -\frac{y}{x} \]

This is a separable differential equation: \[ \frac{dy}{y} = -\frac{dx}{x} \Rightarrow \int \frac{dy}{y} = -\int \frac{dx}{x} \Rightarrow \ln|y| = -\ln|x| + C \Rightarrow \ln|y| + \ln|x| = C \Rightarrow \ln|xy| = C \Rightarrow xy = C' \]

The general solution is: \[ xy = C \]

Use the point \( (3, 2) \) to find the constant: \[ 3 \cdot 2 = 6 \Rightarrow C = 6 \]

Answer:

\[ \boxed{xy = 6} \]