Question

Find the lines that are tangent and normal to the curve at the given point x^2+y^2=25,(3,-4)

Answer 1

That is, the perpendicular slope at (3,-4) is equal to -4/3.

Check via calculus:

\(\large\color{black}{ \displaystyle y=-\sqrt{25-x^2} }\)

\(\large\color{black}{ \displaystyle y'=\frac{x}{\sqrt{25-x^2}} }\)

\(\large\color{black}{ \displaystyle y'=\frac{3}{\sqrt{25-9}} }\)

\(\large\color{black}{ \displaystyle y'=\frac{3}{4} }\)

is the instantaneous slope at (3,-4), and thus the perpendicular slope is correct.