Finding tangents lines passing through a point

Question

How many tangents to $y = x^2 − 1$ pass through the point $(1, −2)$? I found the derivative to be $2x$. Do I plug the $x$ point into $2x$ to get the slope? What do I do next? Do I use a point-slope formula?

Answer 1

the equation through $P(1;2)$ is given by $$y=m(x-1)-2$$ plug this equation into $$y=x^2-1$$ and solve the equation in $x$ and set the discriminat equal to Zero and compute $m$ you have to solve the equation $$m^2-4m-4=0$$

Answer 2

Let a tangent line intersect the parabola at $(a,a^2-1)$. The slope is $2a$. With the point-slope formula, the line equation is $$2a(x-a)=y+1-a^2\\2ax-2a^2=y+1-a^2\\y=2ax-(a^2+1)\text.$$

Let's plug in $(1,-2)$: $$-2=2a-a^2-1\\ \iff a^2-2a-1=0\\ \iff(a-1)^2=2\text.$$

That last equation has 2 real roots, therefore, there are 2 such tangents.

Answer 3

Hint: If a point is “inside” a parabola, there are no tangents to it through the point; if it is on the parabola, there is one; if it is “outside” of the parabola, there are two. This is a consequence of pole-polar relationships: the intersections of the polars of a point with the parabola are where the tangents through the point intersect the parabola.

For this parabola, the interior is the region “above” the curve, i.e., $y\gt x^2-1$.

Answer 4

The equation of tangent line to $y=x^2-1$ at the point $(x_0,y_0)$ is: $$y=y(x_0)+y'(x_0)(x-x_0) \Rightarrow y=x_0^2-1+2x_0(x-x_0).$$ Now the tangent line must pass through $(1,-2)$, so: $$-2=x_0^2-1+2x_0(1-x_0) \Rightarrow x_0^2-2x_0-1=0 \Rightarrow x_0=1\pm \sqrt{2}.$$ It implies there are two tangent lines that touch the parabola at $x_0=1-\sqrt{2}$ and $x_0=1+\sqrt{2}$.