Question

If the parabola $y=(a-b)x^2+(b-c)x+(c-a)$ touches the $x$-axis then the line $ax+by+c=0$

• A. always passes through a fixed point
• B. represents the family of parallel lines
• C. is always perpendicular to $x$-axis
• D. always has negative slope

Answer 1

Answer: (A)

Solving equation of parabola with $x$-axis $(y=0)$, we get
$(a-b)x^2+(b-c)x+(c-a)=0$
which should have two equal values of $x$, as $x$-axis touches the parabola.
$\therefore (b-c{)}^2-4(a-b)(c-a)=0$
$\Rightarrow (b+c-2a{)}^2=0$
$\Rightarrow (b+c-2a{)}^2=0$
$\Rightarrow -2a+b+c=0$
Thus, $ax+by+c=0$ always passes through $(-2,1)$.