Introduction To Calculus Problem (Intersection)

[galeontiger]

Hello.
Right now: we're learning about derivatives.

And the questions reads: Determine the coordinates of two points of tangency to the curve y=-2x^2, given that the corresponding tangent lines intersect at (2,8).

What I know:
I know that the derivative of the function is y'= -4x
I know I have to find two equations of lines (equations of the tangent lines). That when I perform the method of elimination (substition, or subtraction), the results will give me 2 for x and 8 for y.
And maybe there's an easier to find it out using derivatives?

But I'm not sure how to start off this question.
The answer in the back of the book is (2+2root2, -24-16root2) and (2-2root2,-24+16root2).

Thank you for the much needed help.

 

[HallsofIvy]

Hello.
Right now: we're learning about derivatives.

And the questions reads: Determine the coordinates of two points of tangency to the curve y=-2x^2, given that the corresponding tangent lines intersect at (2,8).

What I know:
I know that the derivative of the function is y'= -4x

I know I have to find two equations of lines (equations of the tangent lines). That when I perform the method of elimination (substition, or subtraction), the results will give me 2 for x and 8 for y.
And maybe there's an easier to find it out using derivatives?

But I'm not sure how to start off this question.
The answer in the back of the book is (2+2root2, -24-16root2) and (2-2root2,-24+16root2).

Thank you for the much needed help.

No, there is no easier way- but writing down the equations is not all that difficult.

If a line is tangent to y= -2x² at (x₀, -2x₀²), you know it must pass through the point and have slope -4x₀. That means its equation must be y= (-4x₀)(x- x₀)- 2x₀².

Any such line through (2, 8) must satisfy 8= (-4x₎₀(2- x₀)- 2x₀². That gives you a quadratic to solve for x₀- the two solutions give you the two points.

 

[galeontiger]

Excellent. Thank you. I got the right answer!