Thread: Tangent Lines Perpendicular

find the equation of the line tangent to the curve y=x^4 and perpendicular to 3x-96y=2.

I have thought about it for a while now, and the only part i can think of that might help solve this is that the perpendicular slope is the negative reciprocal of the 3x-96y=2, which would mean the slope is -64.

Any help is much apreciated

Hey DylanRenke.

There is a trick with two-dimensional spaces in that m1*m2 = -1 where m1 is the gradient of one line and m2 is the gradient of the perpendicular line. So if you have an equation y = m1*x + b then a perpendicular line to y will be of the form y_perp = -(1/m1)*x + b2.

So this means the tangent must be equal to -(1/m1) where m1 is the gradient obtained by simplifying 3x-96y=2 into the appropriate form to get m1.

That is exactly what i did to obtain the number -64, but im not sure how to get the 'b' value in the y=mx+b equation so that it is tangent to y=x^4

Re-arranging the formula 3x-96y=2 gives us 96y = 3x - 2 which means y = (3/96)x - 2 in which y = mx + b so m1 = (1/32) and b = -2. This means the perpendicular line will have a gradient of -(1/m1) = -(1/(1/32)) = -32. so any line perpendicular to the first line will have the form y = -32x + b for some y-intercept b.

I think you've calculated the slope of wrong. If you put it in the form , you should get , so then the slope of the perpendicular is -32, not -64.

Now you need to find where the tangent to has slope -32. The slope of the tangent line to a curve is just the derivative (you're learning this in Calculus, right?). If you take the derivative and go through the algebra, you should get . So . So a point on the line is (-2,16), and the slope is -32.

Now plug into the point-slope formula for a line: , so .

I left a few gaps on purpose for you to fill in. Post back if you're still having trouble.

- Hollywood