Re: tangents

Posted by T.Gracken on September 20, 2002 at 07:50:33:

In Reply to: tangents posted by jason on September 19, 2002 at 21:27:49:

: i have to find the equations of all lines tangent to y=9-x² that pass through the point (1,12).
: im stuck on this one.

The first problem I see here is that the point (1,12) is not on the curve...

However, suppose the point (a, b) is on the curve and the tangent line (to the curve) that passes through this point also passes through (1, 12).

Then, since y' = -2x (derivative), the slope of the tangent line will be through (a,b) will be m = -2a.

but also, the slope passing through (a,b) and (1,12) is m = (12-b)/(1-a).

AND!!! since the point (a,b) is on the curve, b = 9-a²

so, -2a = [12-(9-a²)]/(1-a)

Solve this equation for a and you will have the (two) values of a for which the desired line to the curve is tangent

Then use the point slope formula to get your equations...

that is y - 12 = -2a(x - 1)

I' let you try to finish from here,

if you can't, just repost and one of us will get back to you.

Follow Ups:

• Re: tangents Soroban 15:22:53 09/20/02 (2)
  • thank you... T.Gracken 20:44:47 09/20/02 (1)
    • Re: thank you... Joel 13:23:20 09/22/02 (0)

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