Question

Suppose a parabola $y=x^2-ax-1$ intersects the coordinate axes at three points A, B and C respectively. The circumcircle of $△ABC$ intersects the $y$-axis again at the point $D(0,t)$. Find the value of $t$.

Answer 1

Answer: 0001

We have $x=\frac{a\pm \sqrt{a^2+4}}{2}$
$\alpha =\frac{a+\sqrt{a^2+4}}{2};\beta =\frac{a-\sqrt{a^2+4}}{2}$
Equation of family of circles through $A$ and $B$ is
$(x-\alpha )(x-\beta )+y^2+\lambda y=0$
As it passes through $C(0,-1)$,
$\alpha \beta +1-\lambda =0$
$-1+1-\lambda =0\Rightarrow \text{ So }\lambda =0$
(But $\alpha \beta =-1)$
$\therefore$ Equation of circle through $A,B$ and $C$ is $(x-\alpha )(x-\beta )+y^2=0$
It cuts the $y$-axis when $x=0$, so $\alpha \beta +y^2=0$ (Put $\alpha \beta =-1)$
$y^2=1\Rightarrow y=1\text{ or }-1$
Hence $t=1$