Problems from 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 ( $\forall x \in R, f (x) \geq 0$ ) 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?

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Development:

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 $a$ : $f^{'} (x) = 2 x$ $tangency point \equiv (a, a^{2} + 3)$

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: $a^{2} + 3 + 3 = 2 a \cdot (a + 2) \Rightarrow a^{2} + 4 a - 6 = 0 \Rightarrow a = - 2 \pm \sqrt{10}$

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 $(x_{0}, y_{0}) :$

$\begin{array}{rclr} y_{1} = (- 4 + 2 \sqrt{10}) \cdot (x + 2) - 3 & y_{1} = (- 4 + 2 \sqrt{10}) \cdot x - 11 + 4 \sqrt{10} \\ \Rightarrow \\ y_{2} = (- 4 - 2 \sqrt{10}) \cdot (x + 2) - 3 & y_{2} = (- 4 + 2 \sqrt{10}) \cdot x - 11 - 4 \sqrt{10} \end{array}$

Solution:

a) $f (x) = x^{2} + 3$ , $(x_{0}, y_{0}) = (- 2, - 3)$ .

$y_{1} = (- 4 + 2 \sqrt{10}) \cdot x - 11 + 4 \sqrt{10}$

$y_{2} = (- 4 + 2 \sqrt{10}) \cdot x - 11 - 4 \sqrt{10}$

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