Question

Find the equation of the tangent and normal to the curve `x^2+y^2+4x+3y-25=0` at `(-3, 4)`?

Answer 1

Tangent: `11y=2x+50`

Normal: `2y+11x+41=0`

Explanation:

We have `f(x,y)=x^2+y^2+4x+3y-25=0`

In order to differentiate `f`, we must assume that `y=g(x)` and then differentiate `y` using the chain rule. Now, we can actually prove the above assumption, but we don't actually need to.

`f'(x,y)=2x+2y(y')+4+3y'=0`

`-2y(y')-3y'=2x+4`

`y'(-2y-3)=2x+4`

`y'=-(2x+4)/(2y+3)`

`y'(-3,4)=(2(-3)+4)/(2(4)+3)=2/11`

`thereforem_tan=2/11` and `m_normal=-1/(2/11)=-11/2`

Equation of tangent: `y-4=2/11(x+3)`

`y-4=2/11x+6/11`

`y=2/11x+50/11`

`11y=2x+50`

Equation of normal: `y-4=-11/2(x+3)`

`y-4=-11/2x-33/2`

`y+11/2x+25/2=0`

`2y+11x+25=0`

Proof that `y-=g(x)`

`f(x,y)=x^2+4x+y^2+3y-25=0`

Complete the square for both `x` and `y`

`(x+2)^2+(y+3/2)^2-4-9/4-25=0`

`(x+2)^2+(y+3/2)^2=125/4`

`(y+3/2)^2=125/4-(x+2)^2`

`y+3/2=+-sqrt(125/4-(x+2)^2)`

`y=-3/2+-sqrt(125/4-(x+2)^2)`