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

$4(x^2+y^2)\left(2x+2y\dfrac{\mathrm dy}{\mathrm dx}\right)=25\left(2x-2y\dfrac{\mathrm dy}{\mathrm dx}\right)$

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

$40\left(-6+2\dfrac{\mathrm dy}{\mathrm dx}\right)=25\left(-6-2\dfrac{\mathrm dy}{\mathrm dx}\right)\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac9{13}$

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

$y-1=\dfrac9{13}(x+3)\implies\boxed{y=\dfrac{9x+40}{13}}$