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?
Did you sketch that circle and that parabola?
It looks like when x = 8 they have the same slope.
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
To find the tangents common to the given equations, you're on the right track by finding the gradients (or slopes) of the curves. However, you made a small mistake while calculating the gradient of the curve given by the equation y²=16x.
Let's go through the steps to find the gradients correctly:
1. Differentiate the equations with respect to x to find the gradients:
For the equation x² + y² = 8, differentiate both sides with respect to x:
d/dx (x² + y²) = d/dx (8)
2x + 2y * dy/dx = 0
dy/dx = -2x/2y
= -x/y
For the equation y² = 16x, differentiate both sides with respect to x:
d/dx (y²) = d/dx (16x)
2y * dy/dx = 16
dy/dx = 16/2y
= 8/y
2. Now that you have correctly calculated the gradients, equate them to find where they intersect:
-x/y = 8/y
-x = 8
x = -8
Therefore, the point of intersection is (x, y) = (-8, ±√32).
3. To find the corresponding y-values, substitute x = -8 into either of the equations:
For x² + y² = 8:
(-8)² + y² = 8
64 + y² = 8
y² = 8 - 64
y² = -56
Since y² cannot be negative, there are no real solutions for y in this equation.
Similarly, substitute x = -8 into y² = 16x:
y² = 16(-8)
y² = -128
Again, y² cannot be negative, so there are no real solutions for y in this equation either.
4. Since there are no real solutions for y, it means that the two curves do not intersect at any real point, and therefore, there are no common tangents.
Hence, the conclusion is that the equations x² + y² = 8 and y² = 16x do not have any common tangents in the real plane.