Tangents are drawn at points (x_i, y_i); i = | vedclass

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(A) Given points on $y = x^2$ are $(x_i, x_i^2)$.$x_1 = -1$,so $y_1 = 1$.$y_3 = 9 \Rightarrow x_3^2 = 9$. Since $x_1, x_2, x_3$ is an increasing $AP$,

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$4$

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(A) Given points on $y = x^2$ are $(x_i, x_i^2)$.$x_1 = -1$,so $y_1 = 1$.$y_3 = 9 \Rightarrow x_3^2 = 9$. Since $x_1, x_2, x_3$ is an increasing $AP$,$x_3$ must be $3$.Common difference $d = \frac{x_3 - x_1}{3 - 1} = \frac{3 - (-1)}{2} = 2$.Thus,$x_2 = x_1 + d = -1 + 2 = 1$,so $y_2 = 1^2 = 1$.The points are $A(-1, 1)$,$B(1, 1)$,and $C(3, 9)$.The tangents to $y = x^2$ at $(x_i, x_i^2)$ are given by $y - x_i^2 = 2x_i(x - x_i)$,or $y = 2x_ix - x_i^2$.Tangent at $A(-1, 1): y = -2x - 1$.Tangent at $B(1, 1): y = 2x - 1$.Tangent at $C(3, 9): y = 6x - 9$.Intersection of $A$ and $B$: $-2x - 1 = 2x - 1$ $\Rightarrow 4x = 0$ $\Rightarrow x = 0, y = -1$. Point $P(0, -1)$.Intersection of $B$ and $C$: $2x - 1 = 6x - 9$ $\Rightarrow 4x = 8$ $\Rightarrow x = 2, y = 3$. Point $Q(2, 3)$.Intersection of $A$ and $C$: $-2x - 1 = 6x - 9$ $\Rightarrow 8x = 8$ $\Rightarrow x = 1, y = -3$. Point $R(1, -3)$.Area $\Delta = \frac{1}{2} |0(3 - (-3)) + 2(-3 - (-1)) + 1(-1 - 3)| = \frac{1}{2} |0 - 4 - 4| = \frac{1}{2} |-8| = 4$.

Tangents are drawn at points $(x_i, y_i); i = 1, 2, 3$ lying on the parabola $y = x^2$ to enclose a triangle of area $\Delta$. If $x_1, x_2, x_3$ form an increasing arithmetic progression,where $x_1 = -1$ and $y_3 = 9$,then $\Delta$ is ............. $sq. \, units$.

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