Asked by Student
Find the equation of the line tangent to the curve at (5, -3)
(x-2)^2 + (y+3)^2 = 9
I solved the derivative to be dy/dx = ((-2x+4)/ (2y+6))
when i plugged in the points (5, -3) I got the slope as -6/0...How is this possible??
How can i find the equation of this curve if the slope is undefined??
(x-2)^2 + (y+3)^2 = 9
I solved the derivative to be dy/dx = ((-2x+4)/ (2y+6))
when i plugged in the points (5, -3) I got the slope as -6/0...How is this possible??
How can i find the equation of this curve if the slope is undefined??
Answers
Answered by
Reiny
The original equation is the equation of a circle with centre at (2,-3) and a radius of 3
So a tangent at (5,-3) would be a vertical line. Make a rough sketch to see what I mean.
The equation of any vertical line is x = c
in your case
x = 5
So a tangent at (5,-3) would be a vertical line. Make a rough sketch to see what I mean.
The equation of any vertical line is x = c
in your case
x = 5
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.