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

• Aug 3rd 2009, 05:54 PM
potatoes
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?
• Aug 3rd 2009, 06:02 PM
McScruffy
Quote:

Originally Posted by potatoes
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).
• Aug 3rd 2009, 06:04 PM
potatoes
Plug in what x-value? The point (2, -7) is not on the curve, which was my problem.
• Aug 3rd 2009, 06:08 PM
McScruffy
Quote:

Originally Posted by potatoes
Plug in what x-value? The point (2, -7) is not on the curve, which was my problem.

• Aug 3rd 2009, 06:16 PM
songoku
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) ^^
• Aug 3rd 2009, 06:25 PM
VonNemo19
Quote:

Originally Posted by potatoes
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?

$\displaystyle \frac{(2x^2+3)-(-7)}{x-2}=f'(x)$

Solve for x, you must!
• Aug 3rd 2009, 06:50 PM
potatoes
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.
• Aug 3rd 2009, 06:53 PM
songoku
Quote:

Originally Posted by potatoes
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 ^^
• Aug 3rd 2009, 07:07 PM
potatoes
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?
• Aug 3rd 2009, 07:12 PM
VonNemo19
Quote:

Originally Posted by potatoes
Ok so now I tried

4a = 2a^2+7/a-2 (Wondering)

did you see my last post?
• Aug 3rd 2009, 07:18 PM
potatoes
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?
• Aug 3rd 2009, 07:24 PM
VonNemo19
Quote:

Originally Posted by potatoes
Correct?

Indeed.
• Aug 3rd 2009, 07:25 PM
potatoes
Great. Thanks guys!
• Aug 3rd 2009, 07:28 PM
VonNemo19
Quote:

Originally Posted by potatoes
Great. Thanks guys!

(Yes)