# Thread: equation of normal problem

1. ## equation of normal problem

The gradient function of a curve is $\displaystyle kx^2-x,$ where $\displaystyle k$ is constant.The equation of the normal to the curve at point $\displaystyle (1,-2)$ is $\displaystyle 5y+x=7$..

a)Find the value of $\displaystyle k$

b)the equation of the curve.

2. The normal is

$\displaystyle 5y + x = 7$

$\displaystyle y = -\frac{1}{5}x + \frac{7}{5}$.

Therefore, the gradient of the tangent will be $\displaystyle 5$, since the tangent and normal are perpendicular.

So the gradient function at the point $\displaystyle (1, -2)$ has a value of $\displaystyle 5$.

Substituting into the derivative gives

$\displaystyle k(1)^2 - 1 = 5$

$\displaystyle k - 1 = 5$

$\displaystyle k = 6$.

Now that you know $\displaystyle k$, you know

$\displaystyle \frac{dy}{dx} = 6x^2 - x$

$\displaystyle y = \int{6x^2 - x\,dx}$

$\displaystyle y = 2x^3 - \frac{x^2}{2} + C$.

Since $\displaystyle (1, -2)$ lies on this curve

$\displaystyle -2 = 2(1)^3 - \frac{1^2}{2} + C$

$\displaystyle -2 = 2 - \frac{1}{2} + C$

$\displaystyle -\frac{4}{2} = \frac{3}{2} + C$

$\displaystyle -\frac{7}{2} = C$.

Thus the equation of the curve is $\displaystyle y = 2x^3 - \frac{x^2}{2} - \frac{7}{2}$.