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  4. The ends of the latus rectum of the parabola x2 + 10x - 16y + 25 = 0 are

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Q. The ends of the latus rectum of the parabola $x^2 + 10x - 16y + 25 = 0$ are

COMEDKCOMEDK 2009Conic Sections

Solution:

We have, $x^{2} + 10x - 16y + 25 = 0$
$ \Rightarrow \left(x + 5\right)^{2} = 16 y$ ...(i)
$ \Rightarrow X^{2} = 4 \times4y $
Ends of latus rectum of a parabola is $ (\pm \: 2a , a) $
$\Rightarrow X = 2a , X = -2a \Rightarrow x + 5 = 8 , x + 5 = - 8$
$ \Rightarrow x= 3, x = -13$
From (i) when $x = 3, y =4$; when $x =-13, y= 4$
$\therefore $ Required points are (3, 4), (-13, 4)