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

Question

solomonzelman · Answer

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solomonzelman · Answer

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.