Question

Find the equation of tangent and normal line to the given curves..

1.) x^2 + y^2 = 25; (-3,-4)

2.) x^2/32 + y^2/2 =1; (-4,1)

3.) x^2 -6x+2y-8=0; x=0

Answer 1

Same step as the other topic.
You need to solve it in order to "y"
then you need to derivate the given curves
then you need to use this formula: y=f′(a)(x−a)+f(a)
and now you have the equation of the tangent. Something like y=mx+b
To find the normal line you just need to do m=1/m

Answer 2

hmmm maybe not

Answer 3

for 1 take the derivative and get
\[2x+2yy'=0\]
\[y'=-\frac{x}{y}\]

Answer 4

so at
\[(-3,-4)\] the slope is
\[-\frac{3}{4}\] and equation of tangent line is
\[y+4=-\frac{3}{4}(x+3)\]
equation of normal line is
\[y+4=\frac{4}{3}(x+3)\]

Answer 5

second one i cannot read

Answer 6

it means Xsquared over 32 + ysquared over 2 =1

Answer 7

Ellipse is same as circle, implicit diff/normal etc and last is just a normal quadratic.