Let P(a, b) be a point on the parabola y^2 = | vedclass

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(D) The equation of the parabola is $y^2 = 8x$,so $4a = 8 \implies a = 2$. The point $P$ on the parabola is $(2t^2, 2(2)t) = (2t^2, 4t)$.The equation

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$65$

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(D) The equation of the parabola is $y^2 = 8x$,so $4a = 8 \implies a = 2$. The point $P$ on the parabola is $(2t^2, 2(2)t) = (2t^2, 4t)$.The equation of the circle is $x^2 + y^2 - 10x - 14y + 65 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,the centre is $(-g, -f) = (5, 7)$.The tangent to the parabola at $P(2t^2, 4t)$ is given by $ty = x + 2t^2$. Since this tangent passes through $(5, 7)$,we have $7t = 5 + 2t^2$,which simplifies to $2t^2 - 7t + 5 = 0$.Solving for $t$: $(2t - 5)(t - 1) = 0$,so $t = 1$ or $t = 5/2$.For $t = 1$,$P = (2(1)^2, 4(1)) = (2, 4)$.For $t = 5/2$,$P = (2(25/4), 4(5/2)) = (25/2, 10)$.The possible values for $a$ are $2$ and $25/2$. Their product $A = 2 	imes (25/2) = 25$.The possible values for $b$ are $4$ and $10$. Their product $B = 4 	imes 10 = 40$.Therefore,$A + B = 25 + 40 = 65$.

Let $P(a, b)$ be a point on the parabola $y^2 = 8x$ such that the tangent at $P$ passes through the centre of the circle $x^2 + y^2 - 10x - 14y + 65 = 0$. Let $A$ be the product of all possible values of $a$ and $B$ be the product of all possible values of $b$. Then the value of $A + B$ is equal to.

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