# Thread: Equation of Tangent(2) [stucked @ b]

1. ## Equation of Tangent(2) [stucked @ b]

The equation of a curve is $\displaystyle y=2x^2-kx+3$, where k is a constant.

a) Find in terms of k, the gradient of the tangent at point A where x=1.

b) Given that the tangent at point A passes through the point B(5,1), find the value of k.

Attempt

a) $\displaystyle \frac{dy}{dx}= 4x-k$

2. Originally Posted by Punch
The equation of a curve is $\displaystyle y=2x^2-kx+3$, where k is a constant.

a) Find in terms of k, the gradient of the tangent at point A where x=1.

b) Given that the tangent at point A passes through the point B(5,1), find the value of k.

Attempt

a) $\displaystyle \frac{dy}{dx}= 4x-k$
hi

for (a) , you are correct, then substitute x=1 .

(b) it's similar to your previous question . Find the point (A) where the tangent touches the curve in terms of k , then find the gradient of AB which is also the gradient of tangent .

3. I can't find the value of k after finding the gradient in terms of k... please note that the question is asking for the value of k...

4. Originally Posted by Punch
I can't find the value of k after finding the gradient in terms of k... please note that the question is asking for the value of k...
ok , you are not getting what i mean .

point A = (1,5-k) , B (5,1)

so the gradient of AB(tangent) = $\displaystyle \frac{k-4}{4}$ which is also what you found in (1) , 4-k so

$\displaystyle 4-k=\frac{k-4}{4}$

$\displaystyle 16-4k=k-4$

$\displaystyle 5k=20$

$\displaystyle k=4$