If a curve passes through the origin and the slope of the tangent to it at any point $(x, y)$ is $\frac{x^{2}-4 x+y+8}{x-2}$, then this curve also passes through the point :

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Differential Equations

If a curve passes through the origin and the slope of the tangent to it at any point

(x, y)

\frac{x ^{2} - 4 x + y + 8}{x - 2}

, then this curve also passes through the point :

Answer: (A)

Since

x^{2} - 4 x + 8 = (x - 2)^{2} + 4

, the relation is linear in

y

\frac{d y}{d x} - \frac{y}{x - 2} = (x - 2) + \frac{4}{x - 2} .

With integrating factor

\frac{1}{x - 2}

this becomes

\frac{d}{d x} (\frac{y}{x - 2}) = 1 + \frac{4}{( x - 2 ) ^{2}}

, so

\frac{y}{x - 2} = x - \frac{4}{x - 2} + C, y = x (x - 2) - 4 + C (x - 2) .

The curve passes through the origin, so

0 = - 4 - 2 C

and

C = - 2

, which gives

y = x^{2} - 2 x - 4 - 2 x + 4 = x^{2} - 4 x

. Of the points offered only

(5, 5)

satisfies this, since

25 - 20 = 5

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option