If one end of a focal chord AB of the parabola y2=8x is at A12-2, then the equation of the tangent to it at B is :

at12,2at1=12,-24a=8, a=2, 2at1=-2, t1=-12, t1t2=-1, t2=2at12,2at2=(8,8), y2=8x, 2ydydx=8, dydx=82ydydx(8,8)=82(8)=12 ,(y-8)=12(x-8), 2y-16=x-8, -8=x-2y, x-2y+8=0

If one end of a focal chord AB of the parabola y2=8x is at A12-2, then the equation of the tangent to it at B is :

at12,2at1=12,-24a=8, a=2, 2at1=-2, t1=-12, t1t2=-1, t2=2at12,2at2=(8,8), y2=8x, 2ydydx=8, dydx=82ydydx(8,8)=82(8)=12 ,(y-8)=12(x-8), 2y-16=x-8, -8=x-2y, x-2y+8=0

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$If one end of a focal chord AB of the parabola y^{2} = 8 x is at A (\frac{1}{2} - 2), then the equation of$
$the tangent to it at B is :$

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$x + 2 y + 8 = 0$

$2 x + y - 24 = 0$

$x - y + 8 = 0$

$2 x - y - 24 = 0$

answer is C.

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Detailed Solution

$\begin{array}{l} (a t_{1}^{2}, 2 a t_{1}) = (\frac{1}{2}, - 2) \\ 4 a = 8, a = 2, 2 a t_{1} = - 2, t_{1} = - \frac{1}{2}, t_{1} t_{2} = - 1, t_{2} = 2 \\ (a t_{1}^{2}, 2 a t_{2}) = (8, 8), y^{2} = 8 x, 2 y \frac{d y}{d x} = 8, \frac{d y}{d x} = \frac{8}{2 y} \\ \frac{d y}{d x} (8, 8) = \frac{8}{2 (8)} = \frac{1}{2}, (y - 8) = \frac{1}{2} (x - 8), 2 y - 16 = x - 8, - 8 = x - 2 y, x - 2 y + 8 = 0 \end{array}$