Let PQ and RT be two focal chords of the para | vedclass

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(A) The equation of the parabola is $y^2=16x$. Comparing this with $y^2=4ax$,we get $a=4$.Any point on the parabola can be represented as $(at^2, 2at)

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$5$

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(A) The equation of the parabola is $y^2=16x$. Comparing this with $y^2=4ax$,we get $a=4$.Any point on the parabola can be represented as $(at^2, 2at) = (4t^2, 8t)$.For point $P(4,8)$,$4t_1^2=4 \implies t_1=1$ (since $8t_1=8$).For point $R(16,16)$,$4t_2^2=16 \implies t_2=2$ (since $8t_2=16$).Since $PQ$ and $RT$ are focal chords,the product of the parameters of the endpoints of a focal chord is $-1$.For chord $PQ$,$t_P \cdot t_Q = -1 \implies 1 \cdot t_Q = -1 \implies t_Q = -1$. Thus,$Q = (4(-1)^2, 8(-1)) = (4, -8)$.For chord $RT$,$t_R \cdot t_T = -1 \implies 2 \cdot t_T = -1 \implies t_T = -1/2$. Thus,$T = (4(-1/2)^2, 8(-1/2)) = (1, -4)$.The distance $QT$ is given by $\sqrt{(4-1)^2 + (-8 - (-4))^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = \sqrt{25} = 5$.

Let $PQ$ and $RT$ be two focal chords of the parabola $y^2=16x$. If $P=(4,8)$ and $R=(16,16)$,then the length of $QT$ is:

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