Tangents drawn from the point (-8, 0) to the | vedclass

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(A) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which implies $a = 2$. The focus $F$ is $(2, 0)$.The equation of the chord of contact $PQ$

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$48$

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(A) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which implies $a = 2$. The focus $F$ is $(2, 0)$.The equation of the chord of contact $PQ$ for a point $(x_1, y_1) = (-8, 0)$ is given by $T = 0$,which is $yy_1 = 2a(x + x_1)$.Substituting the values,we get $y(0) = 2(2)(x - 8)$,which simplifies to $0 = 4(x - 8)$,so $x = 8$.For $x = 8$,$y^2 = 8(8) = 64$,so $y = \pm 8$. Thus,the points are $P(8, 8)$ and $Q(8, -8)$.The triangle $PFQ$ has vertices $P(8, 8)$,$F(2, 0)$,and $Q(8, -8)$.The base $PQ$ is a vertical line segment with length $|8 - (-8)| = 16$.The height of the triangle from $F(2, 0)$ to the line $x = 8$ is $|8 - 2| = 6$.Area of $	riangle PFQ = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 16 	imes 6 = 48$ sq. units.

Tangents drawn from the point $(-8, 0)$ to the parabola $y^2 = 8x$ touch the parabola at $P$ and $Q$. If $F$ is the focus of the parabola,then the area of the triangle $PFQ$ (in sq. units) is equal to

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