Question

If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is (x² - 4x + y + 8)/(x - 2), then this curve also passes through the point :

• A. (4, 5)
• B. (5, 5)
• C. (5, 4)
• D. (4, 4)

Hint:

Always check for a substitution that turns a complex derivative into a Linear Differential Equation.

Answer 1

The Correct Option is B

Step 1: $\frac{dy}{dx} = \frac{(x-2)^2 + y + 4}{x-2} = (x-2) + \frac{y+4}{x-2}$.
Step 2: Let $Y = y+4$ and $X = x-2$. $\frac{dY}{dX} = X + \frac{Y}{X} \Rightarrow \frac{dY}{dX} - \frac{Y}{X} = X$.
Step 3: Linear D.E.: $I.F. = e^{\int -1/X dX} = 1/X$.
Step 4: $Y(1/X) = \int X(1/X) dX = X + C$.
Step 5: $\frac{y+4}{x-2} = (x-2) + C$. Origin $(0, 0)$ passes through: $\frac{4}{-2} = -2 + C \Rightarrow C = 0$.
Step 6: $y+4 = (x-2)^2 = x^2 - 4x + 4 \Rightarrow y = x^2 - 4x$.
Step 7: At $x=5, y = 25 - 20 = 5$. So point $(5, 5)$ lies on it.