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# Multiple tangents Watch

1. The question asks,

Find all tangents to the curve which go through the point .
This is what I've done so far,

Gradient of tangest when is,

Equation of tangest is therefore,

That's the equation for one of the tangents, but I understand that there's another tangent to the curve, as the point (2,1) is 'below' the curve and as a result, there's a sort of a triangular shape which the two tangents create.

How would I go about finding the other tangent?

Thanks.
2. f'(x) is wrong
3. (Original post by volts)
f'(x) is wrong
Sorry, that was a typo. I've edited it now.
4. The point (1,-2) doesn't lie on f(x). I think you assume it does.
5. (Original post by volts)
The point (1,-2) doesn't lie on f(x). I think you assume it does.
The point lies on the tangents, not on the curve .

So in other wording, it's two tangents to the curve which intersect at the point
6. (Original post by hollywoodbudgie)
The point lies on the tangents, not on the curve .

So in other wording, it's two tangents to the curve which intersect at the point
So why have you evaluated f'(1)?
Also, 5x-7 = f(x) has no real solution
7. (Original post by volts)
So why have you evaluated f'(1)?
Because I don't know how to work this out, hence I made this thread to ask for help.
8. (Original post by hollywoodbudgie)
The point lies on the tangents, not on the curve .

So in other wording, it's two tangents to the curve which intersect at the point
You know that the tangent passes through (1,-2). However, you also know that at the tangent crosses the parabola. So what I would do is find the possible coordinates which lies on the parabola. To do this note that the y coordinate at x1 is x1^+3*x1+3. Now using the formula for the equation of a line with only one point given you get an expression for the possible tangents. You know that the tangents must pass through 1,-2 so substitute these values in for x and y (not x1 and y1). Solve the resulting quadratic equation and you have found the x coordinates which satisfy the conditions given in the question. I'll leave the rest to you
9. (Original post by anshul95)
note that the y coordinate at x1 is x1^+3*x1+3.
Did that equation come out right, I don't particularly understand what the bolded symbol means, '^'. Would it be possible for you to write that out a little clearer please?

Thanks.
10. The gradient of the tangent at a point is given by . So, the equation of a tangent that goes through the point and has gradient is . But you also want the point to lie on the line, and so you must have . Solve this for to find the -values where the tangent of the curve goes through the point. [You have to work out what to use for a,b by the way.]
11. It's 2 tangents which happen to pass through (1, -2). Doesn't mean they touch the curve at this point. They can pass through (1, -2) yet touch the curve elsewhere.

Tangent: y = mx + c; (1, -2) satisfies this
=> -2 = m + c => c = -2 - m
=> y = mx - (m+2)

Since the tangents touch the curve:
mx - (m+2) = x^2 + 3x + 3
=> x^2 + (3 - m)x + (5 + m) = 0

Tangent => Discrim = 0
=> (3 - m)^2 - 4(m + 5) = 0
=> 9 - 6m + m^2 - 4m - 20 = 0
=> m^2 - 10m - 11 = 0
=> (m + 1)(m - 11) = 0
=> m = 11, m = -1

Sub m = 11, m = - 1 into y = mx - (m+2) to get y = 11x - 13 and y = -x - 1.
12. (Original post by Physics Enemy)
It's 2 tangents which happen to pass through (1, -2). Doesn't mean they touch the curve at this point. They can pass through (1, -2) yet touch the curve elsewhere.

Tangent: y = mx + c; (1, -2) satisfies this
=&gt; -2 = m + c =&gt; c = -2 - m
=&gt; y = mx - (m+2)

Since the tangents touch the curve:
mx - (m+2) = x^2 + 3x + 3
=&gt; x^2 + (3 - m)x + (5 + m) = 0

Tangent =&gt; Discrim = 0
=&gt; (3 - m)^2 - 4(m + 5) = 0
=&gt; 9 - 6m + m^2 - 4m - 20 = 0
=&gt; m^2 - 10m - 11 = 0
=&gt; (m + 1)(m - 11) = 0
=&gt; m = 11, m = -1

Sub m = 11, m = - 1 into y = mx - (m+2) to get y = 11x - 13 and y = -x - 1.
Thank you
13. (Original post by hollywoodbudgie)
Thank you
No worries. If it's any consolation, I didn't have a clue what some of the other posters were on about either. Not sure why people make Maths harder than it has to be; it's already hard enough by itself.
14. (Original post by Physics Enemy)
No worries. If it's any consolation, I didn't have a clue what some of the other posters were on about either. Not sure why people make Maths harder than it has to be; it's already hard enough by itself.
This relates to me more than you could possibly imagine.
15. (Original post by hollywoodbudgie)
This relates to me more than you could possibly imagine.
Sorry if we confused you. Some of us just did it another way.
16. (Original post by anshul95)
Sorry if we confused you. Some of us just did it another way.
Thanks for the help.

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