Question

Find the equation of the tangent line to the curve (a lemniscate)

2 (x2 + y2)2 = 25 (x2 - y2) at the point (-3, 1). The equation of this tangentline can be written in the form y = mx + b where m is___________ and b is ___________.

Answer 1

`m=(9)/(13)` and
`b=(40)/(13)`

Explanation:

The equation of curve is

`2(x^2+y^2)^2=25(x^2-y^2)`

We need to find the equation of the tangent line to the curve at the point (-3, 1).

Differentiate with respect to x.

`2[2(x^2+y^2)(d)/(dx)(x^2+y^2)]=25(2x-2y(dy)/(dx))`

`4(x^2+y^2)(2x+2y(dy)/(dx))=25(2x-2y(dy)/(dx))`

The point of tangency is (-3,1). It means the slope of tangent is
`(dy)/(dx)_((-3,1))`.

Substitute x=-3 and y=1 in the above equation.

`4((-3)^2+(1)^2)(2(-3)+2(1)(dy)/(dx))=25(2(-3)-2(1)(dy)/(dx))`

`40(-6+2(dy)/(dx))=25(-6-2(dy)/(dx))`

`-240+80(dy)/(dx))=-150-50(dy)/(dx)`

`80(dy)/(dx)+50(dy)/(dx)=-150+240`

`130(dy)/(dx)=90`

Divide both sides by 130.

`(dy)/(dx)=(9)/(13)`

If a line passes through a points
`(x_1,y_1)` with slope m, then the point slope form of the line is

`y-y_1=m(x-x_1)`

The slope of tangent line is
`(9)/(13)` and it passes through the point (-3,1). So, the equation of tangent is

`y-1=(9)/(13)(x-(-3))`

`y-1=(9)/(13)(x)+(27)/(13)`

Add 1 on both sides.

`y=(9)/(13)(x)+(27)/(13)+1`

`y=(9)/(13)(x)+(40)/(13)`

Therefore,
`m=(9)/(13)` and
`b=(40)/(13)`.