Tangent straight line to a curve at a point
a) Define a parable $f(x)$ with all its points in the first and second quadrant ($\forall {x} \in \mathbb{R} , f(x)\geq 0$) and a point $(x_0,y_0)$ placed in the third or fourth quadrant ($y_0<0$).
b) Find it the tangent straight lines to $f(x)$ which goes through that point. How many tangents can you find?
a) We could use $f(x)=x^2+3$, and the point $(x_0,y_0)=(-2,-3)$.
b) First, we use the expression of a general straight line going through the point $(x_0,y_0):$ $$y-y_0=m\cdot(x-x_0)$$ $$y+3=m\cdot (x+2)$$
Then we take the derivative of $f(x)$ and we obtain a generic tangency point parameterized by $a$: $$f'(x)=2x$$ $$\mbox{tangency point}\equiv(a,a^2+3)$$
The slope of the tangent straight line will be given by $m=2\cdot a$, and that straight line will go through the touching point $(a,a^2+3)$. Substituting in the equation for the tangent straight line: $$a^2+3+3=2a\cdot (a+2) \Rightarrow a^2+4a-6=0 \Rightarrow a=-2\pm\sqrt{10}$$
We should note that there are to possible values for the slope $(m=2\cdot a):$ $$m_1=-4+2\sqrt{10}$$ $$m_2=-4-2\sqrt{10}$$
This means that there will be two tangent straight lines to $f(x)$ and that they go through $(x_0,y_0):$
$$\begin{eqnarray} & y_1=(-4+2\sqrt{10})\cdot(x+2)-3 & & y_1=(-4+2\sqrt{10})\cdot x-11+4\sqrt{10} \\\\ & & \Rightarrow & \\\\ & y_2=(-4-2\sqrt{10})\cdot(x+2)-3 & & y_2=(-4+2\sqrt{10})\cdot x-11-4\sqrt{10} \end{eqnarray}$$
a) $f(x)=x^2+3$, $(x_0,y_0)=(-2,-3)$.
b)
$y_1=(-4+2\sqrt{10})\cdot x-11+4\sqrt{10}$
$$y_2=(-4+2\sqrt{10})\cdot x-11-4\sqrt{10}$$