Question

The family of curves whose $x$ and $y$ intercepts of a tangent at any point are respectively double the $x$ and $y$ coordinates of that point is:

• A. $xy = C$
• B. $x^2 + y^2 = C$
• C. $x^2 - y^2 = C$
• D. $\frac{y}{x} = C$

Answer 1

The Correct Option is A

1. Understand the condition:

The tangent at any point \( (x, y) \) has \( x \)-intercept \( 2x \) and \( y \)-intercept \( 2y \).

2. Find the equation of the tangent:

The tangent line passes through \( (2x, 0) \) and \( (0, 2y) \). Its slope is:

\[ m = \frac{0 - 2y}{2x - 0} = -\frac{y}{x} \]

Equation of the tangent:

\[ Y - y = -\frac{y}{x}(X - x) \]

3. Find the differential equation:

At \( X = 0 \), \( Y = 2y \):

\[ 2y - y = -\frac{y}{x}(-x) \implies y = y \]

This holds true, so we use the slope condition:

\[ \frac{dy}{dx} = -\frac{y}{x} \]

4. Solve the differential equation:

Separate variables:

\[ \frac{dy}{y} = -\frac{dx}{x} \implies \ln y = -\ln x + C \implies \ln(xy) = C \implies xy = e^C = C' \]

Correct Answer: (A) \( xy = C \)

Answer 2

For a curve $xy = C$, consider a point $(a, b)$ on the curve. 

The equation of the tangent at this point is derived using implicit differentiation: \[ b \, dx + a \, dy = 0 \implies \frac{dy}{dx} = -\frac{b}{a}. \] The equation of the tangent line is: \[ y - b = -\frac{b}{a}(x - a) \implies y = -\frac{b}{a}x + 2b. \] The $x$-intercept is when $y = 0$: \[ 0 = -\frac{b}{a}x + 2b \implies x = 2a. \] The $y$-intercept is when $x = 0$: \[ y = 2b. \] Thus, the intercepts are $(2a, 0)$ and $(0, 2b)$, which are double the coordinates of the point $(a, b)$. 

Therefore, the family of curves satisfying the given condition is $xy = C$.