Question

The graph of the function `(x^2 + y^2 + 12x + 9)^2 = 4(2x + 3)^3` is a Tricuspoid as shown below. (a) Find the point on the curve, above x-axis, with x = 0? (b) Find slope of tangent line to the point in part (a)?

Answer 1

(a) `(1.18,0)` and (b) `y=-0.68125x+1.18`

Explanation:

When `x=0`, `(x^2+y^2+12x+9)^2=4(2x+3)^3` reduces to

`(y^2+9)^2=108` or `y^4+18y^2-27=0`

and `y^2=(-18+-sqrt(18^2-4xx1xx(-27)))/2`

= `(-18+-sqrt432)/2=-9+-6sqrt3`

and ignoring `-9-6sqrt3` as it is negative

`y=+-sqrt(6sqrt3-9)`

(a) and above `x=0` the point is `(sqrt(6sqrt3-9),0)` or `(1.18,0)`

(b) For slope of tangent we should find the first derivative of function `f(x,y)=0`. It is

`2(x^2+y^2+12x+9)(2x+2y(dy)/(dx)+12)=12(2x+3)^2xx2`

and when `x=0` and `y=sqrt(6sqrt3-9)`, this becomes

`2(6sqrt3-9+9)(2sqrt(6sqrt3-9)(dy)/(dx)+12)=216`

or `24sqrt3sqrt(6sqrt3-9)(dy)/(dx)=216-144sqrt3`

and `(dy)/(dx)=(216-144sqrt3)/(24sqrt3sqrt(6sqrt3-9))=(3sqrt3-6)/sqrt(6sqrt3-9)`

= `-0.68125`

and equation of tangent is `y=-0.68125x+1.18`

graph{(y+0.68125x-1.18)((x^2+y^2+12x+9)^2-4(2x+3)^3)=0 [-5, 5, -2.5, 2.5]}