Question

Find an equation of the tangent line to the curve 2(x2+y2)2=25(x2−y2) (a lemniscate) at the point (−3,1). An equation of the tangent line to the lemniscate at the given point is

Answer 1

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

Let
`y=y(x)`, so that differentiating both sides wrt
`x` gives

`4(x^2+y^2)\left(2x+2y(\mathrm dy)/(\mathrm dx)\right)=25\left(2x-2y(\mathrm dy)/(\mathrm dx)\right)`

If
`x=-3` and
`y=1`, the above reduces to

`40\left(-6+2(\mathrm dy)/(\mathrm dx)\right)=25\left(-6-2(\mathrm dy)/(\mathrm dx)\right)\implies(\mathrm dy)/(\mathrm dx)=\frac9{13}`

This is the slope of the tangent line, which has equation

`y-1=\frac9{13}(x+3)\implies\boxed{y=(9x+40)/(13)}`