# Math Help - equation of normal problem

1. ## equation of normal problem

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

a)Find the value of $k$

b)the equation of the curve.

2. The normal is

$5y + x = 7$

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

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

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

Substituting into the derivative gives

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

$k - 1 = 5$

$k = 6$.

Now that you know $k$, you know

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

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

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

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

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

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

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

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

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