The axis of a parabola is parallel to the Y-a | vedclass

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(D) Since the points $(1,0), (0,2),$ and $(-1,-1)$ lie on the parabola $ax^2 + bx + cy + d = 0$,we have:$a(1)^2 + b(1) + c(0) + d = 0 \Rightarrow a +

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$10$

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(D) Since the points $(1,0), (0,2),$ and $(-1,-1)$ lie on the parabola $ax^2 + bx + cy + d = 0$,we have:$a(1)^2 + b(1) + c(0) + d = 0 \Rightarrow a + b + d = 0$ ... $(i)$$a(0)^2 + b(0) + c(2) + d = 0 \Rightarrow 2c + d = 0$ ... $(ii)$$a(-1)^2 + b(-1) + c(-1) + d = 0 \Rightarrow a - b - c + d = 0$ ... $(iii)$Subtracting $(iii)$ from $(i)$ gives: $(a + b + d) - (a - b - c + d) = 0$ $\Rightarrow 2b + c = 0$ $\Rightarrow c = -2b$.From $(ii)$,$d = -2c = -2(-2b) = 4b$.From $(i)$,$a = -b - d = -b - 4b = -5b$.Thus,$\frac{ad}{bc} = \frac{(-5b)(4b)}{b(-2b)} = \frac{-20b^2}{-2b^2} = 10$.

The axis of a parabola is parallel to the $Y$-axis. If this parabola passes through the points $(1,0), (0,2), (-1,-1)$ and its equation is $ax^2 + bx + cy + d = 0$,then $\frac{ad}{bc} = $

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