# Tangents

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#1
I really need help with this question anyone please:

10. The tangent to the curve y = x3 − 3x2 − 8x + 5 at point A goes through point C(–13, –3). The gradient of curve at point A is p.

(i) Find the equation of the line through A and C in terms of p.

(ii) Use differentiation to obtain another equation involving p.

(iii) Hence by eliminating p and y, find the cubic equation in x that represents the possible positions of the x-coordinate of A.

(iv) Given further that the x-coordinate of A is rational, use your calculator to find the coordinates of A and the value of p.

(v) The line perpendicular to AC through C is also the normal to the curve at point B. Find point B and hence the area of triangle ABC.
1
3 months ago
#2

Do you have any thoughts so far or are you completely stuck?
Last edited by ElMoro; 3 months ago
0
#3
Well for part (a) I did y1-y2 = p(x1-x2), which would get: y--3=p(x--13) then y + 3=p(x + 13) then y + 3=px + 13x then y=px + 13x-3 ?
Last edited by Sbrown_2005; 3 months ago
0
3 months ago
#4
Nice!

So to elaborate on the steps you've taken:

You took the information that "the gradient of curve at point A is " and also realised that the gradient of the curve at point A must be equal to the gradient of the tangent to the curve (at point A).

Therefore, the line through A and C (which is the tangent to the curve at point A) must have gradient .

You also knew that the equation of a straight line (in 2D) can have the form , where is the gradient of the curve and is some point on the line.

You correctly plugged in the gradient and point - which has coordinates - into this equation to get: You made an error when expanding the brackets on the right-hand side, though. Can you see your mistake?
0
#5
(Original post by ElMoro)
Nice!

So to elaborate on the steps you've taken:

You took the information that "the gradient of curve at point A is " and also realised that the gradient of the curve at point A must be equal to the gradient of the tangent to the curve (at point A).

Therefore, the line through A and C (which is the tangent to the curve at point A) must have gradient .

You also knew that the equation of a straight line (in 2D) can have the form , where is the gradient of the curve and is some point on the line.

You correctly plugged in the gradient and point - which has coordinates - into this equation to get: You made an error when expanding the brackets on the right-hand side, though. Can you see your mistake?
I genuinely can't see where I've gone wrong
0
3 months ago
#6
(Original post by Sbrown_2005)
I genuinely can't see where I've gone wrong
Maybe it was just a typo? But when you expand the brackets of You would get (Notice that both terms on the RHS have a p in them. You had written px + 13x above)

In any case, you can subtract 3 from both sides and you'd have the equation you're looking for. Well done! You did that all yourself, really.

What are your thoughts for part (ii)?
Last edited by ElMoro; 3 months ago
0
#7
(Original post by ElMoro)
Maybe it was just a typo? But when you expand the brackets of You would get (Notice that both terms on the RHS have a p in them. You had written px + 13x above)

In any case, you can subtract 3 from both sides and you'd have the equation you're looking for. Well done! You did that all yourself, really.

What are your thoughts for part (ii)?
I differentiated it to get 3x2-6x-8, which would be p, then I substituted that back into the other p equation would should then hopefully get y= 3x^3+33x^2-86x-112. Then when I made that equal to 0, I got 3 x-values which were, (1)x=-12.9861071 (2)x=2.958002034 (3)x=-0.9718949366. Then what...….
Last edited by Sbrown_2005; 3 months ago
0
3 months ago
#8
(Original post by Sbrown_2005)
I differentiated it to get 3x2-6x-8, which would be p, then I substituted that back into the other p equation would should then hopefully get y= 3x^3+33x^2-86x-112. Then when I made that equal to 0, I got 3 x-values which were, (1)x=-12.9861071 (2)x=2.958002034 (3)x=-0.9718949366. Then what...….
I think the bold is the equation of the tangent, though you should check the constant.

Rather than setting it equal to zero though (find the roots of the tangent) as you have done, you want to equate it to the original curve to find the points of intersection of the curve and the tangent. In other words the tangent point, A, lies on both the curve and the tangent line.
Last edited by mqb2766; 3 months ago
0
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