Question

The slope of the tangent to the curve at a point $(x,y)$ on it is proportional to $(x-2)$. If the slope of the tangent to the curve at $(10,9)$ on it is $3$. The equation of the curve is:

• A. $y=k(x-2{)}^2$
• B. $y=\frac{-3}{16}(x-2{)}^2+1$
• C. $y=\frac{3}{8}(\frac{x^2}{2}-2x)-3$
• D. $y=K(x+2{)}^2$

Answer 1

Answer: (C)

$y=\frac{3}{8}(\frac{x^2}{2}-2x)-3$

Given that the slope of the curve is proportional to $(x-2)$

Let the curve be $y=f\left(x\right)$

$\therefore$ the slope of the curve at a point is$\frac{d}{dx}f\left(x\right)$,

$\frac{d}{dx}f\left(x\right)=k(x-2)$

$k$ is the proportionality constant.

Given that the slope of the curve at the point $(10,9)$ on it is $3$

$⟹3=k(10-2)$

$⟹k=\frac{3}{8}$

Now,

$\frac{d}{dx}f\left(x\right)=\frac{3}{8}(x-2)$

Integrating the equation with respect to $x$ on both sides gives,

$f\left(x\right)=\frac{3x^2}{16}-\frac{3x}{4}+c$

Where $c$ is the integration constant,As the curve passes through $(10,9)$,

$9=\frac{300}{16}-\frac{30}{4}+c$

$\therefore c=-\frac{9}{4}$

$\therefore f\left(x\right)=\frac{3}{8}(\frac{x^2}{2}-2x)-\frac{9}{4}$