# Thread: normal line to the curve

1. ## normal line to the curve

A curve $y=f(x)$ with the gradient function $\frac{dy}{dx}=\frac{x^2}{3x-1}.$

The straight line $ky=2x-7$ is normal to the curve $y=f(x)$ at $(2,m)$.Find the value of $k$ and $m$.

any help will appreciate,.

2. Originally Posted by mastermin346
The straight line $ky=2x-7$ is normal to the curve $y=f(x)$ at $(2,m)$.Find the value of $k$ and $m$.

any help will appreciate,.
There is not enough information given. Go back and check the question.

3. Originally Posted by mr fantastic
There is not enough information given. Go back and check the question.

hi sir..

sorry,my mistake

the question is. A curve $y=f(x)$ with gradient function $\frac{dy}{dx}=\frac{x^2}{3x-1}$.

4. Well start by substituting $x=2$ into $\frac{dy}{dx}$ what you you get? What does that mean?

5. Originally Posted by pickslides
Well start by substituting $x=2$ into $\frac{dy}{dx}$ what you you get? What does that mean?
hi,i get $\frac{dy}{dx}=\frac{4}{5}$ then?it mean the gradient right?

6. Originally Posted by mastermin346
hi,i get $\frac{dy}{dx}=\frac{4}{5}$ then?it mean the gradient right?
Correct, it is the gradient of $f(x)$ at $x=2$. Now what is the relationship between the gradient and the normal?

Hint: $m_N\times m_T = -1$

How can we use this value?

7. Originally Posted by pickslides
Correct, it is the gradient of $f(x)$ at $x=2$. Now what is the relationship between the gradient and the normal?

Hint: $m_N\times m_T = -1$

How can we use this value?
i get the gradient of normal is $-\frac{5}{4}$.

8. Originally Posted by mastermin346

The straight line $ky=2x-7$ is normal to the curve $y=f(x)$ at $(2,m)$.
so maybe its' time to find $k$

Making the form $y=mx+c$

$ky=2x-7\implies y=\frac{2}{k}x-\frac{7}{k}$

$\frac{-5}{4}= \frac{2}{k}$