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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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
2 Answers
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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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Answer
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