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Math Help - Find equation of tangent line that passes through point not on curve?

  1. #1
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    Find equation of tangent line that passes through point not on curve?

    The question is: Find the equations of the tangent lines to the curve y = 2x^2 + 3 That pass through the point (2, -7)

    The last time I did this sort of questions was over a year ago and I think I remember that you're supposed to pick a point (a, f(a) ) on the parabola first, and go from there. But what then? How would you find the slope from (2, -7) to a point you don't know?
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    Quote Originally Posted by potatoes View Post
    The question is: Find the equations of the tangent lines to the curve y = 2x^2 + 3 That pass through the point (2, -7)

    The last time I did this sort of questions was over a year ago and I think I remember that you're supposed to pick a point (a, f(a) ) on the parabola first, and go from there. But what then? How would you find the slope from (2, -7) to a point you don't know?
    Take the derivative, plug in your x-value to get the slope at the point of tangency. Then find the equation for the line, using your slope and point (I'd use the point-slope form of the equation for a line).
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  3. #3
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    Plug in what x-value? The point (2, -7) is not on the curve, which was my problem.
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    Member McScruffy's Avatar
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    Quote Originally Posted by potatoes View Post
    Plug in what x-value? The point (2, -7) is not on the curve, which was my problem.
    Sorry, I misread your question.
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  5. #5
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    Hi potatoes

    Ok, you have picked point (a, f(a) ) on the parabola. So, of course you can find the slope in term of a. You also can find the coordinate where the parabola and tangent meet, in term of a.

    Find the tangent (still consists of a), then do something with the tangent and (2, -7) ^^
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    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by potatoes View Post
    The question is: Find the equations of the tangent lines to the curve y = 2x^2 + 3 That pass through the point (2, -7)

    The last time I did this sort of questions was over a year ago and I think I remember that you're supposed to pick a point (a, f(a) ) on the parabola first, and go from there. But what then? How would you find the slope from (2, -7) to a point you don't know?
    \frac{(2x^2+3)-(-7)}{x-2}=f'(x)

    Solve for x, you must!
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  7. #7
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    Hmm I gave it a go but I think I'm doing it wrong.

    So the slope of the tangent is 4a

    and the x co-ord of the tangent on the line is a, and the y co-ord is 2a^2+3.

    So I made the tangent y+7=4a(x-2) --> 2a^2+3+7=4a(a-2)

    and I solved that and I got 2 a values: (-4+sqrt(8))/2 and (-4-sqrt(8))/2

    So then I took the first a value and calculated f(a) = 3.68

    So now I did f(a)-y/a-x ---> 3.68+7/-0.585-2 = -4.13 (slope)

    Tangent is y+7 = -4.13(x-2) ?

    When I put this tangent into the graphing program I have it cuts the curve at 2 points (close together though) so is this not the tangent? Can someone tell me how to do it lol.
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  8. #8
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    Quote Originally Posted by potatoes View Post
    So I made the tangent y+7=4a(x-2) --> 2a^2+3+7=4a(a-2)
    You said it by yourself that (2,-7) is not on the parabola ^^
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  9. #9
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    Ok so now I tried

    4a = 2a^2+7/a-2

    and I got a=4 and a=1

    With a = 1 the slope of the tangent would be 4a which is 4

    so y-5=4(x-1) for one of the tangents?

    Is this right?
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by potatoes View Post
    Ok so now I tried

    4a = 2a^2+7/a-2
    did you see my last post?
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  11. #11
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    ooops I solved that last equation wrong. I did it again and I got a = 5 and a = 1 for 4a = 2a^2+3+7/a-2

    and the tangents are y-5=-4(x+1) and y-53=20(x-5) . Correct?
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  12. #12
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by potatoes View Post
    Correct?
    Indeed.
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  13. #13
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    Great. Thanks guys!
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  14. #14
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by potatoes View Post
    Great. Thanks guys!
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