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 point:

As per questions d y d x = x 2 − 4 x + y + 8 x − 2 d y d x = ( x − 2 ) 2 + ( y + 4 ) ( x − 2 ) d y d x = ( x − 2 ) + y + 4 x − 2 ......(i) Let y + 4 x − 2 = t (y + 4) = t(x – 2) Putting in equation (i) ( x − 2 ) d t d x + t = ( x − 2 ) + t d t d x = 1 dt = dx Integrating on both the sides t = x + c y + 4 x − 2 = x + c Passing through origin C = -2 ∴ equation of curve y + 4 x − 2 = x − 2

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:

Option 1 - (5, 5)

Option 2 - (4, 4)

Option 3 - (4, 5)

Option 4 - (5, 4)

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Answered by

Alok Kumar Singh | Contributor-Level 10

7 months ago

Correct Option - 1

Detailed Solution:

As per questions

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

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

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

Let $\frac{y + 4}{x - 2} = t$

(y + 4) = t(x – 2)

Putting in equation (i)

$(x - 2) \frac{d t}{d x} + t = (x - 2) + t$

$\frac{d t}{d x} = 1$

dt = dx

Integrating on both the sides t = x + c

$\frac{y + 4}{x - 2} = x + c$

Passing through origin C = -2

$∴$ equation of curve $\frac{y + 4}{x - 2} = x - 2$

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