Find all values of 'a' such that y=a/(x−9) and y=x^2−18x+81 intersect at right angles.

show steps please :)

I did this one on Wednesday

http://www.jiskha.com/display.cgi?id=1319657997

thanks!

To find the values of 'a' such that the two curves intersect at right angles, we need to find the condition for the slopes of the curves to be negative reciprocals of each other.

Let's start by finding the derivatives of the two curves:

For y = a/(x-9), we need to rewrite it as y = a(x-9)^(-1).
Then, we can differentiate y with respect to x to find its slope:
dy/dx = -a(x-9)^(-2) * 1

For y = x^2 - 18x + 81, we can differentiate it directly to find its slope:
dy/dx = 2x - 18

Now, for the two curves to intersect at right angles, the slopes of the curves must be negative reciprocals of each other. So we have:

- a(x-9)^(-2) * 1 * (2x - 18) = -1

Simplifying the equation, we get:

2ax - 18a(x-9)^(-2) = -1

To continue solving for 'a', we need more information.

To find the values of 'a' such that the two curves intersect at right angles, we need to find the condition for the product of the slopes of the two curves to be -1.

We start by finding the slopes of the two curves. The slope of the curve y = a/(x−9) can be found by differentiating the equation with respect to x:

dy/dx = -a / (x-9)^2

The slope of the curve y = x^2 - 18x + 81 can also be found by differentiating the equation with respect to x:

dy/dx = 2x - 18

Now, we equate the two slopes and solve for x:

- a / (x-9)^2 = 2x - 18

Multiply both sides by (x-9)^2 to eliminate the denominator:

- a = (2x - 18) * (x-9)^2

Expanding the right side of the equation:

- a = (2x - 18) * (x^2 - 18x + 81)

Simplifying further:

- a = 2x^3 - 36x^2 + 162x - 18x^2 + 324x - 1458

Combining like terms:

- a = 2x^3 - 54x^2 + 486x - 1458

Now, we differentiate this equation with respect to x to find the slope of the curve y = - a / (x-9)^2:

d(-a)/dx = 6x^2 - 108x + 486

For the two curves to intersect at right angles, the product of the slopes must be -1. So we have:

(-a) * (6x^2 - 108x + 486) = -1

Expanding and rearranging the equation:

6ax^2 - 108ax + 486a + 1 = 0

This is a quadratic equation in terms of 'x'. For a unique real solution, the discriminant (b^2 - 4ac) must be equal to zero. So we have:

(-108a)^2 - 4 * 6a * (486a + 1) = 0

Simplifying:

11664a^2 - 11664a^2 - 24a = 0

Subtracting like terms and dividing by -24a:

-24a = 0

Dividing both sides by -24:

a = 0

Therefore, the only value of 'a' for which the two curves intersect at right angles is '0'.