Let the point P of the focal chord PQ of the | vedclass

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(A) The equation of the parabola is $y^2=16x$. Comparing this with $y^2=4ax$,we get $a=4$. The focus $S$ is $(a, 0) = (4, 0)$.Let the coordinates of p

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$17$

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(A) The equation of the parabola is $y^2=16x$. Comparing this with $y^2=4ax$,we get $a=4$. The focus $S$ is $(a, 0) = (4, 0)$.Let the coordinates of point $P$ be $(at_1^2, 2at_1)$. Given $P = (1, -4)$,we have $2at_1 = -4$ $\Rightarrow 2(4)t_1 = -4$ $\Rightarrow t_1 = -\frac{1}{2}$.Since $PQ$ is a focal chord,the product of the parameters of its endpoints is $t_1 t_2 = -1$. Thus,$t_2 = -\frac{1}{t_1} = -\frac{1}{-1/2} = 2$.The coordinates of point $Q$ are $(at_2^2, 2at_2) = (4(2)^2, 2(4)(2)) = (16, 16)$.Let the focus $S(4, 0)$ divide the chord $PQ$ in the ratio $m:n$. Using the section formula for the $x$-coordinate:$4 = \frac{m(x_Q) + n(x_P)}{m+n} = \frac{m(16) + n(1)}{m+n}$$4(m+n) = 16m + n$$4m + 4n = 16m + n$$3n = 12m \Rightarrow \frac{m}{n} = \frac{3}{12} = \frac{1}{4}$.Since $\operatorname{gcd}(m, n) = \operatorname{gcd}(1, 4) = 1$,we have $m=1$ and $n=4$.Therefore,$m^2+n^2 = 1^2+4^2 = 1+16 = 17$.

Let the point $P$ of the focal chord $PQ$ of the parabola $y^2=16x$ be $(1, -4)$. If the focus of the parabola divides the chord $PQ$ in the ratio $m:n$,where $\operatorname{gcd}(m, n)=1$,then $m^2+n^2$ is equal to:

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