Find equation of tangent line knowing hyperbola and point on line

Question

I have a problem I've been trying to solve, but I was not able to do it.

A hyperbola is $x^2-y^2=16$ and a point is $(-1,-7)$, not on the curve.

Find equation of tangent line to the hyperbola passing through that point.

Answer 1

Using implicit differentiation, $2x +2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y} = -\frac{1}{7}$.

Now we have the slope of the tangent line i.e. $y = -\frac{1}{7} +c$ to find $c$ we substitute the point $(-1,-7)$ into out equation as this point lies on the tangent line.

$(-7) = \frac{1}{7}+c \Rightarrow c = -\frac{50}{7}$ so the equation of the tangent is $y = -\frac{1}{7}x-\frac{50}{7}$

Answer 2

$$x^2+y^2=6 \Rightarrow 2x+2y \frac{dy}{dx}=0$$

So $$\frac{dy}{dx}=-\frac{x}{y} \Rightarrow \frac{dy}{dx}=-\frac{1}{7}$$

$$\frac{y-(-7)}{x-(-1)}=-\frac{1}{7}$$

$$7y+49=-x-1$$

$$7y+x+50=0$$

Answer 3

From geometry: the tangent to a circle is perpendicular to the radius at that point.

The center of the circle is $(0,0)$ so the gradient of the radius to point $(-1,-7)$ is $$m_r=\frac{0-(-7)}{0-(-1)}=7$$

From the perpendicular property we have that the gradient of the tangent is $$m_t=\frac{-1}{m_r}$$ $$m_t=-\frac{1}{7}$$

Now we have $$y=-\frac{1}{7}x+c$$ and since the point $(-1,-7)$ is on the tangent line we can substitute to find $c$. $$-7=-\frac{1}{7}(-1)+c$$ $$c=\frac{-50}{7}$$ $$y=-\frac{1}{7}x-\frac{50}{7}$$

Answer 4

Choose $y$ for an independent variable, so that your hyperbola is simply described as two curves, one branch with an equation $$x=+\sqrt{y^2+4^2}$$ and the other one $$x=-\sqrt{y^2+4^2}$$

Then you can calculate a slope $\frac{dx}{dy}$ of a tangent line for each branch, depending on the tangency point's $y$ coordinate, find a tangent line equation and check if it is satisfied by the given point's coordinates — just like Daniel, Angelo Mark and Pieter Rousseau show in their answers.

For the given point and curve you should get one tangent line for each branch.