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 :

Solution

Maths sir · 4 steps

Split the numerator around x - 2, and the slope equation becomes a linear differential equation in y.

1. Split the numerator. Write x² - 4x + 8 = (x-2)² + 4. Then the slope is dy/dx = (x-2) + 4/(x-2) + y/(x-2).
2. Linear form. Move the y term to the left: dy/dx - 1/(x-2) y = (x-2) + 4/(x-2). The integrating factor is e⁻ˡⁿ⁽ˣ⁻²⁾ = 1/(x-2). Multiplying by it gives d/dx(y/(x-2)) = 1 + 4/(x-2)².

   y/(x-2) = x - 4/(x-2) + C ⇒ y = x(x-2) - 4 + C(x-2)

3. Use the origin. Put x = 0, y = 0: 0 = 0 - 4 - 2C, so C = -2. Then y = x² - 2x - 4 - 2x + 4, that is, y = x² - 4x.

   Point	x² - 4x	Matches y?
   (4, 4)	16 - 16 = 0	no
   (5, 5)	25 - 20 = 5	yes
   (5, 4)	25 - 20 = 5	no
   (4, 5)	16 - 16 = 0	no

4. So. The curve is y = x² - 4x, and it passes through (5, 5), which is option (b).

The idea to keep. When the slope has y appearing only to the first power, rearrange it into dy/dx + P(x) y = Q(x) and use the integrating factor. Splitting awkward numerators first keeps the integration clean.

Answer from the NTA final answer key. Solution by PrepWiser.