First slide

Parabola

Question

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:

Moderate

A

32
B

$16 \sqrt{2}$
C

$16 \sqrt{3}$
D

$32 \sqrt{2}$

Solution

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

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