Parabola, secant line, perpendicular and distance (II)

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Given the parabola $y=x^2+2x-3$:

• (a) find the vertex and the axis intersections;
• (b) draw it;
• (c) find its intersections with the line Pina through $A(1;2)$ and $B(2;-3)$;
• (d) find the line Gina perpendicular to Pina through $A$;
• (e) find the distance from $B$ to Pina.

Solution

(a) Vertex: $x_V=-1$, $y_V=-4$, i.e. $V(-1;-4)$. $\cap x$: $(x+3)(x-1)=0\Rightarrow x=-3,\,1$; $\cap y$: $(0;-3)$. (c) Pina through $A(1;2)$ and $B(2;-3)$: slope $-5$, so $y=-5x+7$. Intersections with the parabola: $x^2+7x-10=0\Rightarrow x=\dfrac{-7\pm\sqrt{89}}{2}\approx1{,}22\ \text{and}\ -8{,}22$. (d) Pina has slope $-5$; Gina (perpendicular) has slope $\dfrac15$ through $A(1;2)$: $y=\dfrac15 x+\dfrac95$. (e) Pina passes through $B$, so the distance of $B$ from Pina is $d=\frac{|-10+3+7|}{\sqrt{26}}=0.$ $\boxed{V(-1;-4);\ \text{Gina}: y=\tfrac15 x+\tfrac95;\ d(B,\text{Pina})=0\ (B\in\text{Pina}).}$