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Write an equation for the tangent line to hyperbola x^2-4y^2=12 at point (-4,1)?
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You need to remember the form of equation of the tangent line to the graph of a given function `f(x)` , at a given point `(x_p,y_p)` , such that:
`f(x) - y_p = f'(x_p)(x - x_p)`
Let `f(x) = y` , such that:
`y - y_p = f'(x_p)(x - x_p)`
Since the problem provides the coordinates of the point ` (-4,1)` , yields:
`y - 1 = f'(-4)(x + 4)`
You need to evaluate the derivative of the function and you need to calculate `f'(-4)` , such that:
`x^2-4y^2=12 => 4y^2 = x^2 - 12 => y^2 = (x/2)^2 - 3`
`y = +-sqrt((x/2)^2 - 3)`
`y = f(x) => f'(x) = (((x/2)^2 - 3)')/(2((x/2)^2 - 3))`
`f'(x)= x/(4((x/2)^2 - 3)) => f'(-4) = -4/(4*sqrt(4 - 3)) => f'(-4) = -1`
Replacing `-1` for `f'(-4)` in equation of tangent line yields:
`y - 1 = -(x + 4) => y = -x - 4 + 1=> y = -x - 3`
Hence, evaluating the equation of the tangent line to hyperbola `x^2 - 4y^2 = 12` , at the point `(-4,1)` , yields `y = -x - 3.`
Posted by sciencesolve on September 18, 2013 at 3:42 PM (Answer #1)
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