If (2, -8) is one endpoint of a focal chord o | vedclass

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(A) For a parabola $y^{2} = 4ax$,if one endpoint of a focal chord is $(at_{1}^{2}, 2at_{1})$,then the other endpoint is $(at_{2}^{2}, 2at_{2})$ where

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$(32, 32)$

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(A) For a parabola $y^{2} = 4ax$,if one endpoint of a focal chord is $(at_{1}^{2}, 2at_{1})$,then the other endpoint is $(at_{2}^{2}, 2at_{2})$ where $t_{1}t_{2} = -1$.Given $y^{2} = 32x$,we have $4a = 32$,so $a = 8$.The point $(2, -8)$ corresponds to $(at_{1}^{2}, 2at_{1})$.$2at_{1} = -8 \implies 2(8)t_{1} = -8 \implies 16t_{1} = -8 \implies t_{1} = -\frac{1}{2}$.Since $t_{1}t_{2} = -1$,we have $t_{2} = -\frac{1}{t_{1}} = -\frac{1}{-1/2} = 2$.The other endpoint is $(at_{2}^{2}, 2at_{2}) = (8(2)^{2}, 2(8)(2)) = (8 	imes 4, 32) = (32, 32)$.

If $(2, -8)$ is one endpoint of a focal chord of the parabola $y^{2} = 32x$,what is the other endpoint?

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