Pencil through a point and pencil with a fixed base point

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

• (a) draw the parabola;
• (b) write, in terms of a parameter $m$, the equation of the generic line through $P(-4;4)$;
• (c) find the values of $m$ for which the line is tangent, external or secant to the parabola;
• (d) consider the lines $ay-x+2=0$ as $a$ varies: show that they all pass through a single point and find it;
• (e) show that these lines can never be tangent to the parabola.

Solution

The drawing of the parabola (vertex $V(4;4)$, opening downward) is left to the student.

(b) Proper pencil with base point $P(-4;4)$: $y-4=m(x+4)$, i.e. $y=mx+4m+4$.

(c) Intersection with the parabola: $-\dfrac{x^2}{8}+x+2=mx+4m+4$. Multiplying by $-8$ and rearranging: $x^2+(8m-8)x+(32m+16)=0, \qquad \Delta=64m^2-256m=64m(m-4).$

• tangent $\Delta=0 \implies m=0 \ \vee\ m=4$;
• secant $\Delta>0 \implies m<0 \ \vee\ m>4$;
• external $\Delta<0 \implies 0<m<4$.

(d) The equation $ay-x+2=0$ can be written $a\cdot y+(-x+2)=0$: it is a pencil whose lines all pass through the point common to the “generators” $y=0$ and $-x+2=0$, i.e. $x=2$, $y=0$. The base point is $P(2;0)$.

(e) For $a\neq 0$, $y=\dfrac{x-2}{a}$; substituting into the parabola and ordering in $x$: $a x^2+(8-8a)x-16-16a=0, \qquad \Delta=64\big(2a^2-a+1\big).$ The trinomial $2a^2-a+1$ has $\Delta'=1-8<0$ and is always positive, so $\Delta>0$ for every $a\neq 0$: the lines are always secant, never tangent. (The line $a=0$ is the vertical $x=2$, also not tangent.)

$\boxed{\text{tangent: } m=0 \vee m=4;\quad \text{fixed base point } P(2;0);\quad \text{never tangent}}$