If the tangent drawn at the point P(4,8) to t | vedclass

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(B) The equation of the tangent to the parabola $y^2=16x$ at point $P(4,8)$ is given by $8y = 8(x+4)$,which simplifies to $y = x+4$ $\dots(i)$.Since t

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$(4,8)$

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(B) The equation of the tangent to the parabola $y^2=16x$ at point $P(4,8)$ is given by $8y = 8(x+4)$,which simplifies to $y = x+4$ $\dots(i)$.Since the tangent $(i)$ intersects the parabola $y^2 = 16x+80$ at points $A$ and $B$,we substitute $y = x+4$ into the second equation:$(x+4)^2 = 16x + 80$$x^2 + 8x + 16 = 16x + 80$$x^2 - 8x - 64 = 0$.Let the coordinates of $A$ be $(x_1, y_1)$ and $B$ be $(x_2, y_2)$. The roots of the quadratic equation $x^2 - 8x - 64 = 0$ are $x_1$ and $x_2$. The sum of the roots is $x_1 + x_2 = 8$.The mid-point $M(h, k)$ of $AB$ has coordinates $h = \frac{x_1+x_2}{2} = \frac{8}{2} = 4$.Since the mid-point lies on the line $y = x+4$,we have $k = h+4 = 4+4 = 8$.Thus,the mid-point of $AB$ is $(4,8)$.

List-$I$	List-$II$
$(I)$ Vertex	$(A)$ $(-\frac{3}{2}, -3)$
$(II)$ Focus	$(B)$ $(\frac{3}{2}, -3)$
$(III)$ Equation of the directrix	$(C)$ $2x+5=0$
$(IV)$ Equation of the axis	$(D)$ $2x+y+3=0$
	$(E)$ $y+3=0$
	$(F)$ $(-2, -3)$

If the tangent drawn at the point $P(4,8)$ to the parabola $y^2=16x$ meets the parabola $y^2=16x+80$ at $A$ and $B$,then the mid-point of $AB$ is

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