Let y2 = 16x be a given parabola and L be an extremity of its latus rectum in the first quadrant. If a chord is drawn through L with slope-1, then the length of this chord is:

Equation of the latus rectum is $x = 4$ and the coordinates of L are $(4, 8)$ . Equation of the chord through L with slope-1

is $y - 8 = - (x - 4)$

$\Rightarrow x + y = 12$

Solving the equation of the chord and the parabola we get

$y^{2} = 16 (12 - y) \Rightarrow y = 8$ and -24

So the coordinates of M the other end of the chord through L is $(36, - 24) .$

and $L M = \sqrt{(36 - 4)^{2} + (- 24 - 8)^{2}} = 32 \sqrt{2}$

Let $y^{2} = 16 x$ be a given parabola and L be an extremity of its latus rectum in the first quadrant. If a chord is drawn through L with slope-1, then the length of this chord is: