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Q.
The circle drawn on the latus rectum of the parabola $4y
^2 + 25 = 4 (y + 4 x)$as diameter cuts the axis of
the parabola at the points :
Conic Sections
Solution:
Given equation after dividing by 4
$y^{ 2 }- y- 4x +\frac{25}{4}=0 $
$ \left(y-\frac{1}{2}\right)^{2}=4\left(x-\frac{3}{2}\right) $
$ y^{2}=4X where y-\frac{1}{2}=Y$
$ x-\frac{3}{2}=X$
Equation of a circle with $AB$ as diametre and $XY$ axis
$(X-1)^2 + Y^2 - 4 = 0 $
put Y = 0 to get
$X^2 - 2X - 3 = 0 $
$X = 3 or X = - 1 $
Hence $P(3,0)$ and $Q(-1,0) $w.r.t $XY$
$\therefore P\left(\frac{9}{2}\frac{1}{2}\right)$ and $Q \left(\frac{1 }{2 }\frac{1 }{2 }\right)$ w.r.t. $xy$
