Let P(alpha, beta) be a point on the parabola | vedclass

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(C) The equation of the chord of the parabola $x^2 = 8y$ with midpoint $(x_1, y_1) = (1, 5/4)$ is given by $T = S_1$.Here,$T = x x_1 - 4(y + y_1)$ and

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$192$

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(C) The equation of the chord of the parabola $x^2 = 8y$ with midpoint $(x_1, y_1) = (1, 5/4)$ is given by $T = S_1$.Here,$T = x x_1 - 4(y + y_1)$ and $S_1 = x_1^2 - 8y_1$.Substituting the values,we get $x(1) - 4(y + 5/4) = 1^2 - 8(5/4)$.$x - 4y - 5 = 1 - 10$.$x - 4y - 5 = -9 \Rightarrow x - 4y + 4 = 0$.Since $P(\alpha, \beta)$ lies on this chord,$\alpha - 4\beta + 4 = 0$,which implies $\alpha = 4\beta - 4$.Also,$P(\alpha, \beta)$ lies on $y^2 = 4x$,so $\beta^2 = 4\alpha$.Substituting $\alpha$ in the second equation: $\beta^2 = 4(4\beta - 4) = 16\beta - 16$.$\beta^2 - 16\beta + 16 = 0$.This gives $\beta = \frac{16 \pm \sqrt{256 - 64}}{2} = 8 \pm \sqrt{48} = 8 \pm 4\sqrt{3}$.Since $\alpha = 4\beta - 4$,we have $\alpha - 28 = 4\beta - 4 - 28 = 4\beta - 32 = 4(\beta - 8)$.Thus,$(\alpha - 28)(\beta - 8) = 4(\beta - 8)^2$.Since $(\beta - 8) = \pm 4\sqrt{3}$,then $(\beta - 8)^2 = 48$.Therefore,$(\alpha - 28)(\beta - 8) = 4 	imes 48 = 192$.

Let $P(\alpha, \beta)$ be a point on the parabola $y^2 = 4x$. If $P$ also lies on the chord of the parabola $x^2 = 8y$ whose midpoint is $(1, 5/4)$,then $(\alpha - 28)(\beta - 8)$ is equal to:

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