Question

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

• A. $(4,5)$
• B. $(5,4)$
• C. $(4,4)$
• D. $(5,5)$

Answer 1

The correct option is D

$(5,5)$

Explanation for the correct option:

Step 1. Find the required point:

The slope will be:

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

$\Rightarrow$ $\frac{dy}{dx}=\frac{{\left(x-2\right)}^2+y+4}{x-2}$

$\Rightarrow$ $\frac{dy}{dx}=\left(x-2\right)+\frac{y}{x-2}+\frac{4}{x-2}$

$\Rightarrow$$\frac{dy}{dx}-\frac{y}{x-2}=\left(x-2\right)+\frac{4}{x-2}$

$\Rightarrow$I.F. $=e^{-\int \frac{1}{x-2}dx}$

$=\frac{1}{x-2}$

Step 2. Solution of L.D.E is given by:

$y\left(\frac{1}{x-2}\right)=\int \frac{1}{x-2}\left(\left(x-2\right)+\frac{4}{x-2}\right)dx$

$\Rightarrow$$\frac{y}{x-2}=\int \left(1\right)dx+\int \frac{4}{{\left(x-2\right)}^2}dx$

$\Rightarrow$$\frac{y}{x-2}=x-\frac{4}{x-2}+C$ …(1)

Now, at $x=0,y=0$

$\frac{0}{0-2}=0-\frac{4}{0-2}+C$

$\Rightarrow$ $C=-2$

Step 3. Put the value of $C$ in equation (1), we get

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

$\Rightarrow$ $y=x(x-2)-4-2(x-2)$

$\Rightarrow$ $y=x^2-4x$ …(2)

Step 4. Check all the given points if they are satisfy equation (2)

(A) $(4,5)$

$y=x^2-4x$

$\Rightarrow$$5=4^2-4\times 4$

$\Rightarrow$$5=16-16$

$\Rightarrow$$5\ne 0$

(B) $(5,4)$

$y=x^2-4x$

$\Rightarrow$$4=5^2-4\times 5$

$\Rightarrow$$4=25-20$

$\Rightarrow$$4\ne 5$

(C) $(4,4)$

$y=x^2-4x$

$\Rightarrow$$4=4^2-4\times 4$

$\Rightarrow$$4=16-16$

$\Rightarrow$$4\ne 0$

(D) $(5,5)$

$y=x^2-4x$

$\Rightarrow$$5=5^2-4\times 5$

$\Rightarrow$$5=25-20$

$\Rightarrow$$5=5$

Thus, The satisfying point of this equation is $\left(5,5\right)$

Hence, Option ‘D’ is Correct.