If a curve passes through the origin and the slope of the tangent to it at any point x - y is x 2-4 x - y -8- x -2- then this curve also passes through the pointA- 5-4B- 4-4- 5-5D- 4-5

Dydx=(x 2)^2+y+4(x 2)=(x 2)+y+4(x 2) Let x 2=t⇒ dx=dt and y+4=u⇒ dy=du nbsp; dydx=dudt nbsp; dudt=t+ut⇒dudt ut=t nbsp; I.F.=e^∫ 1tdt=e^ ln t=1t nbsp ...

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Byju's Answer

Standard XII

Mathematics

Conic Section

If a curve pa...

Question

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 :

(4, 5)

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(5, 4)

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(4, 4)

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(5, 5)

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Solution

The correct option is D $(5, 5)$
$\frac{d y}{d x} = \frac{(x - 2)^{2} + y + 4}{(x - 2)} = (x - 2) + \frac{y + 4}{(x - 2)}$
Let $x - 2 = t \Rightarrow d x = d t$
and $y + 4 = u \Rightarrow d y = d u \frac{d y}{d x} = \frac{d u}{d t} \frac{d u}{d t} = t + \frac{u}{t} \Rightarrow \frac{d u}{d t} - \frac{u}{t} = t I . F . = e^{\int \frac{- 1}{t} d t} = e^{- ln t} = \frac{1}{t} u \times \frac{1}{t} = \int t \times \frac{1}{t} d t \Rightarrow \frac{u}{t} = t + c \frac{y + 4}{x - 2} = (x - 2) + c$
Passing through $(0, 0)$
$c = 0$
$\Rightarrow (y + 4) = (x - 2)^{2}$
It passes though $(5, 5)$

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

The correct option is D (5,5)dydx=(x−2)2+y+4(x−2)=(x−2)+y+4(x−2) Let x−2=t⇒dx=dt and y+4=u⇒dy=dudydx=dudtdudt=t+ut⇒dudt−ut=tI.F.=e∫−1tdt=e−lnt=1tu×1t=∫t×1tdt⇒ut=t+cy+4x−2=(x−2)+c Passing through (0,0) c=0 ⇒(y+4)=(x−2)2 It passes though (5,5)

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 :