Tangent lines

[Kaikai28]

How to find the equation of the tangents to the hyperbola xy=1 through the point (-1,1) ??

 

[Romsek]

a) write the hyperbola in the form [MATH]y = f(x)[/MATH]
b) the slope of the tangent line at [MATH](-1,1)[/MATH] will be [MATH]\left .\dfrac{df}{dx} \right |_{x=-1}[/MATH]
c) you now have the slope and a point on the line. Use the point slope formula to derive the equation of the line.

 

[lookagain]

a) write the hyperbola in the form [MATH]y = f(x)[/MATH]
b) the slope of the tangent line at [MATH](-1,1)[/MATH] will be [MATH]\left .\dfrac{df}{dx} \right |_{x=-1}[/MATH]
c) you now have the slope and a point on the line. Use the point slope formula to derive the equation of the line.

(-1, 1) does not lie on xy = 1. There are tangent lines from the point (-1, 1) to the
hyperbola xy = 1. This is a different problem that requires additional/some different
steps.

 

[Romsek]

(-1, 1) does not lie on xy = 1. There are tangent lines from the point (-1, 1) to the
hyperbola xy = 1. This is a different problem that requires additional/some different
steps.

yep. Ignore the previous post.

Ok.. There exist a point on each of the branches of the hyperbola such that the line from that point to the point (-1,1) is tangent
to the hyperbola at that point.

Looking at the point [MATH](p_x, p_y)[/MATH] in the second quadrant we have

[MATH]\dfrac{f(p_x) - 1}{p_x - (-1)} = f^\prime(p_x)[/MATH]
[MATH]f(x) = \dfrac 1 x\\ f^\prime(x) = -\dfrac{1}{x^2}[/MATH]
[MATH] \dfrac{\frac{1}{p_x}-1}{p_x+1} = -\dfrac{1}{p_x^2}\\ p_x - p_x^2 = -p_x - 1\\ p_x^2 - 2p_x - 1 = 0\\ p_x = 1+\sqrt{2} \text{ (the other solution is not in the second quadrant)}\\ p_y = f(p_x) = \dfrac{1}{1+\sqrt{2}} \text{so our point in the second quadrant is $\left(1+\sqrt{2},~\dfrac{1}{1+\sqrt{2}}\right)$} [/MATH]
I leave you to solve for the point on the other branch of the hyperbola.

 

[jonah2.0]

Beer soaked contribution follows.
Alternatively, using algebra, the equation of the line passing through (-1,1) is given by y-1=m[x-(-1)] or y=mx+m+1, where m is the slope. Substitute y into the equation for the hyperbola and solve for x using the quadratic formula. Take the discriminant and set it to zero and solve for m.