Question

The circle drawn on the latus rectum of the parabola $4y^2+25=4(y+4x)$as diameter cuts the axis of the parabola at the points :

• A. $\left(\frac{1}{2},\frac{1}{2}\right)\left(\frac{9}{2},\frac{1}{2}\right)$
• B. $\left(\frac{1}{2},\frac{1}{2}\right)\left(\frac{1}{2},\frac{9}{2}\right)$
• C. $\left(\frac{1}{2},\frac{1}{2}\right)\left(0,0\right)$
• D. $\left(\frac{1}{2},\frac{7}{2}\right)\left(-\frac{1}{2},\frac{9}{2}\right)$

Answer 1

Answer: (A)

Given equation after dividing by 4
$y^2-y-4x+\frac{25}{4}=0$
${\left(y-\frac{1}{2}\right)}^2=4\left(x-\frac{3}{2}\right)$
$y^2=4Xwherey-\frac{1}{2}=Y$
$x-\frac{3}{2}=X$
Equation of a circle with $AB$ as diametre and $XY$ axis
$(X-1{)}^2+Y^2-4=0$
put Y = 0 to get
$X^2-2X-3=0$
$X=3orX=-1$
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. $xy$