Did you sketch that circle and that parabola?
It looks like when x = 8 they have the same slope.
I'm studying analytic geometry and i came accross this question;"find the tangents common to x²+ y²=8 and y²=16x.How am i going to work it out,please?
I tried to work out their gradients which gave me
x²+y²=8
2x+2ydy/dx=0
dy/dx=-2x/2y
=-x/y
y²=16
2ydy/dx=16
dy/dx=8/y
so my question is,am i wrong and if not how am i going to proceed?
2 answers
At a point (h,4√h) the parabola has slope 2/√h
At a point (k,√(8-k^2)) the circle has slope -k/√(8-k^2)
(Assume top branch of both curves.)
We want the two slopes to be the same, and we want the slope to be ∆y/∆x, so
2/√h = -k/√(8-k^2)
2/√h = (4√h-√(8-k^2))/(h-k)
Slosh that around for a while and you find h=4, k=-2, so the points
(-2,2) on the circle and
(4,8) on the parabola
are on the common tangent.
Damon was right, with a lot less work. But to verify his assertion, you'd still have to have found the point on the circle with slope=1
At a point (k,√(8-k^2)) the circle has slope -k/√(8-k^2)
(Assume top branch of both curves.)
We want the two slopes to be the same, and we want the slope to be ∆y/∆x, so
2/√h = -k/√(8-k^2)
2/√h = (4√h-√(8-k^2))/(h-k)
Slosh that around for a while and you find h=4, k=-2, so the points
(-2,2) on the circle and
(4,8) on the parabola
are on the common tangent.
Damon was right, with a lot less work. But to verify his assertion, you'd still have to have found the point on the circle with slope=1
1 answer