# Find the equation of the tangent line to a given curve at a given point

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?

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Oh,I forgot to write that I found that x=-5 by replacing the given data in the line. –  egdfd Nov 28 '12 at 23:36
If we put $x=-5$, $y=3$, I think we get $y'=-4/3$. –  André Nicolas Nov 28 '12 at 23:41

When $x=-5$ and $y=3$, we get $y'=-\dfrac{4}{3}$.

Everything else is correct, so the equation of the line only needs a minor fix. It is possible that an answer of the shape $y=mx+b$ is expected.

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i) By replacing the value of y by 3 we get two values of x = -5 and x = 3. since we are looking for a value of x which less than 0 then In this case x = -5 would be the choice.

ii) Differentiate the equation ; 2ydy = 2x + x -- dx iii) then replacing X == -5 in above or y' = 4/3

iv) Equation of tangent in the form y2-y1 = m( x2 - x1)

                                 3-y   =   4/3( -5 -x)
y   =  3 - 4/3 ( -5-x)


v) hope this will help

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