Find the equation of the tangent line to the curve $x^2 - y^2 +2x-6=0$ in the point $(x,3)$, where $x<0.$ So I tried to find the derivative of the given curve, $2x-2yy' +2=0$...here I replaced the given coordinates and I have that $y'=-3/2$ I replace in $y-3=-1.5(x+5)$ and thats it...is this correct?
Find the equation of the tangent line to a given curve at a given point
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derivatives
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0Oh,I forgot to write that I found that x=-5 by replacing the given data in the line. – 2012-11-28
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2If we put $x=-5$, $y=3$, I think we get $y'=-4/3$. – 2012-11-28
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0It's not quite clear to me... Does it say first to find such value (or values) of $x$ that $(x,3)$ belongs to a hyperbola and then find an equation of a tangent line in that point? Or may be they request to find a line (or lines) tangent to the hyperbola passing the $(x,3)$ point for any possible $x$ for which such tangents exists (as a function of the $x$, of course)? – 2015-10-07