Distances and intersections between a line and a parabola

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Consider the point $M(-1;3)$, the line $d:2x+y-4=0$ and the parabola $y=\dfrac{x^2}{2}+3x+4$:

• (a) find the distance between $M$ and the line $d$;
• (b) find the distance between $M$ and the vertex of the parabola;
• (c) find the intersections between $d$ and the parabola.

Solution

(a) Point-to-line distance. With $d:2x+y-4=0$ (i.e. $A=2$, $B=1$, $C=-4$) and $M(-1;3)$: $d(M,d)=\frac{|2(-1)+3-4|}{\sqrt{2^2+1^2}}=\frac{|-3|}{\sqrt5}=\frac{3}{\sqrt5}=\frac{3\sqrt5}{5}\approx 1.342$

(b) Distance from the vertex. The vertex of the parabola has $x_V=-\dfrac{b}{2a}=-\dfrac{3}{2\cdot\frac12}=-3$ and $y_V=\dfrac{(-3)^2}{2}+3(-3)+4=-\dfrac12$, so $V\left(-3;-\dfrac12\right)$. $d(M,V)=\sqrt{(-1+3)^2+\left(3+\tfrac12\right)^2}=\sqrt{4+\tfrac{49}{4}}=\sqrt{\tfrac{65}{4}}=\frac{\sqrt{65}}{2}\approx 4.031$

(c) Line-parabola intersections. From the line $d$: $y=-2x+4$. Substituting into the parabola: $-2x+4=\frac{x^2}{2}+3x+4 \;\Longrightarrow\; 0=\frac{x^2}{2}+5x \;\Longrightarrow\; x\left(\frac{x}{2}+5\right)=0$ hence $x=0$ (then $y=4$) and $x=-10$ (then $y=-2(-10)+4=24$). The points are $(0;4)$ and $(-10;24)$. $\boxed{d(M,d)=\tfrac{3\sqrt5}{5}\approx1.342\quad d(M,V)=\tfrac{\sqrt{65}}{2}\approx4.031\quad (0;4),\ (-10;24)}$