Pencil of lines and tangency condition to a parabola

Text

Given the parabola $y=-\dfrac{x^2}{8}+x$:

• (a) draw the parabola;
• (b) write, in terms of a parameter $m$, the equation of the generic line through $P(-3;2)$;
• (c) find the values of $m$ for which the line is tangent, external or secant to the parabola;
• (d) write, in terms of a parameter $q$, the equation of the generic line parallel to $y=x$;
• (e) find the values of $q$ for which the line is tangent, external or secant to the parabola.

Solution

The parabola has vertex $V(4;2)$ and opens downward (the drawing is left to the student).

(b) Proper pencil with base point $P(-3;2)$: $y-2=m(x+3)$, i.e. $y=mx+3m+2$.

(c) Intersection with the parabola: $-\dfrac{x^2}{8}+x=mx+3m+2$. Multiplying by $-8$ and rearranging: $x^2+(8m-8)x+(24m+16)=0, \qquad \Delta=(8m-8)^2-4(24m+16)=64m^2-224m=32m(2m-7).$

• tangent $\Delta=0 \implies m=0 \ \vee\ m=\dfrac72$;
• secant $\Delta>0 \implies m<0 \ \vee\ m>\dfrac72$;
• external $\Delta<0 \implies 0<m<\dfrac72$.

(d) Improper pencil of lines parallel to $y=x$ (slope $1$): $y=x+q$.

(e) Intersection: $-\dfrac{x^2}{8}+x=x+q \implies x^2=-8q$.

• tangent a single solution $\implies q=0$;
• secant two solutions $\implies q<0$;
• external no solution $\implies q>0$.

$\boxed{\text{tangent: } m=0 \vee m=\tfrac72;\quad q=0}$