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I was given the following problem:
"Consider the curve defined by $y^2-x^2y=6$ for $y>0$"
"Write an equation for the line tangent to the curve at the point $(1,3)$"
So I thought I could approach it like I would if I were finding the line tangent to a point on a regular curve, thus I evaluated the derivative at the point and plugged that value into the equation for point-slope (and ended up getting an answer of: $y-3=(x-1)$) but I have no clue if this is correct. Any help would be appreciated!
hint
$ y$ beeing $>0$ we have
$$y=\frac{x^2+\sqrt{x^4+24}}{2}$$ $$=f(x)=g(x^2)$$
with
$$g(X)=\frac 12\Bigl(X+\sqrt{24+X^2}\Bigr)$$
the tangent line equation at the point $(1,3)$ is
$$\color{red}{y}=f'(1)(x-1)+f(1)$$ $$=2g'(1)(x-1)+3$$
where $$2g'(x)=1+\frac{x}{\sqrt{x^2+24}}$$
Si, $$f'(1)=1+\frac 15=\frac 65$$ finally $$\color{red}{y}=\frac 65x+\frac 95$$
