Question bank/JEE Main/Mathematics

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

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$\ We\ have\ x = \frac{a \pm \sqrt{a^{2} + 4}}{2}$
$\alpha = \frac{a + \sqrt{a^{2} + 4}}{2};\beta = \frac{a - \sqrt{a^{2} + 4}}{2}$
[IMAGE] Equation of family of circles through
$A\ and\ B\ is\ (x - \alpha)(x - \beta) + y^{2} + \lambda y = 0$
As it passes through
$C(0, - 1),\alpha\beta + 1 - \lambda = 0(\ But\ \alpha\beta = - 1)$
-1+1-$\lambda = 0 \Rightarrow \lambda = 0$
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
$\alpha\beta + y^{2} = 0\ (Put\ \left. \ \alpha\beta = - 1 \right)$
y^2=1 $\Rightarrow$ y=1 or -1
Hence t=1

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx