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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(5, 4)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(4, 4)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(5, 5)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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 $(5, 5)$

Suggest Corrections

Similar questions

Q. 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 :

Q. If a curve passes through the point

(1, - 2)

and has slope of the tangent at any point

(x, y)

on it as

\frac{x^{2} - 2 y}{x}

, then the curve also passes through the point :

Q. If a curve passes through the point

(1, - 2)

and has slope of the tangent at any point

(x, y)

on it as

\frac{x^{2} - 2 y}{x}

, then the curve also passes through the point :

Q. If the curve

y = a x^{2} + b x + c, x \in R,

passes through the point

(1, 2)

and the tangent line to this curve at origin is

y = x,

then the possible values of

a, b, c

are

Q. Given that the slope of the tangent to a curve

y = y (x)

at any point

(x, y)

\frac{2 y}{x^{2}}

. If the curve passes through the centre of the circle

x^{2} + y^{2} - 2 x - 2 y = 0

, then its equation is :

Join BYJU'S Learning Program

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 :