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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!

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  • $\begingroup$ You have to show us how you evaluated the derivative so we can answer if it is correct. $\endgroup$
    – Miguel
    Commented Sep 13, 2020 at 23:13

1 Answer 1

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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 $$\color{red}{y}=\frac 65x+\frac 95$$

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