Line-parabola intersection and whether a point belongs to a curve

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Given the line $R:3x+2y-3=0$, the parabola $L:y=\dfrac{x^2}{3}+x-3$, the parabola $G:y=-\dfrac{x^2}{4}+4$ and the point $P(3;-2)$:

• (a) find the intersection between $L$ and $R$;
• (b) find the distance between $R$ and $P$;
• (c) decide whether $P$ belongs to $L$, to $G$, or to neither.

Solution

(a) Intersection $L\cap R$. From the line: $y=\dfrac{3-3x}{2}$. Set it equal to $L$ and multiply by $6$: $\frac{3-3x}{2}=\frac{x^2}{3}+x-3 \;\Longrightarrow\; 9-9x=2x^2+6x-18 \;\Longrightarrow\; 2x^2+15x-27=0$ $x=\frac{-15\pm\sqrt{225+216}}{4}=\frac{-15\pm\sqrt{441}}{4}=\frac{-15\pm21}{4}$ hence $x=\dfrac{6}{4}=\dfrac32$ (then $y=\dfrac{3-4.5}{2}=-0.75$) and $x=-9$ (then $y=\dfrac{3+27}{2}=15$). The points are $\left(\tfrac32;-\tfrac34\right)$ and $(-9;15)$.

(b) Distance $R$–$P$. With $R:3x+2y-3=0$ and $P(3;-2)$: $d(R,P)=\frac{|3\cdot3+2(-2)-3|}{\sqrt{3^2+2^2}}=\frac{|9-4-3|}{\sqrt{13}}=\frac{2}{\sqrt{13}}=\frac{2\sqrt{13}}{13}\approx 0.555$

(c) Does $P(3;-2)$ belong to a curve? $L$: $\dfrac{3^2}{3}+3-3=3\neq-2$; $G$: $-\dfrac{3^2}{4}+4=1.75\neq-2$. So $P$ belongs to neither $L$ nor $G$. $\boxed{L\cap R:\left(\tfrac32;-\tfrac34\right),\ (-9;15)\quad d(R,P)=\tfrac{2\sqrt{13}}{13}\approx0.555\quad P\notin L,\ P\notin G}$