Parabola, secant line, perpendicular and distance (I)

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Given the parabola $y=x^2+5x+6$:

• (a) find the vertex and the axis intersections;
• (b) draw it;
• (c) find its intersections with the line through $A(2;2)$ and $B(4;-3)$;
• (d) find the line Gina perpendicular to the line Pina $y=-2x+3$, through $A$;
• (e) find the distance from $B$ to Pina.

Solution

(a) Vertex: $x_V=-2{,}5$, $y_V=-0{,}25$, i.e. $V(-2{,}5;-0{,}25)$. $\cap x$: $(x+2)(x+3)=0\Rightarrow x=-2,\,-3$; $\cap y$: $(0;6)$. (c) Line $AB$: slope $-\dfrac52$, so $y=-\dfrac52 x+7$. Intersections: $2x^2+15x-2=0\Rightarrow x=\dfrac{-15\pm\sqrt{241}}{4}\approx0{,}13\ \text{and}\ -7{,}63$. (d) Pina has slope $-2$; the perpendicular Gina has slope $\dfrac12$ through $A(2;2)$: $y=\dfrac12 x+1$. (e) Pina: $2x+y-3=0$. Distance of $B(4;-3)$: $d=\frac{|8-3-3|}{\sqrt5}=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}\approx0{,}89.$ $\boxed{V(-2{,}5;-0{,}25);\ \text{Gina}: y=\tfrac12 x+1;\ d(B,\text{Pina})=\tfrac{2\sqrt5}{5}\approx0{,}89.}$