Question

Given below are Matching type questions, with two columns(each having some items) each.Each item of column$I$ has to be matched with the items of column$II$, by enclosing the correct match.
Note$:$An item of column$I$ can be matched with more than one items of column$II$.All the items of column $II$ have to be matched.

Answer 1

(A)$\because y^2-4x-2y+5=0$ is arranged as
$y^2-2y+1-1-4x+5=0$
$\Rightarrow y^2-2y+1=4x-4$
$\Rightarrow {(y-1)}^2=4(x-1)$
Let $y-1=Y$ and $x-1=X$
$\therefore Y^2=4X$
On comparing with $Y^2=4aX$ we get $a=1$
According to the quaestion $X>2a$
or $x-1>2$ for $a=1$
or $x>2+1=3$
$\therefore x$ takes the values $>3$
$\therefore x=4,5,6,7,8$
(B)$\because 4y^2-32x+4y+65=0$
$\Rightarrow 4(y^2+y)-32x+65=0$
$\Rightarrow 4(y^2+y+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2)-32x+65=0$
$\Rightarrow 4{(y+\frac{1}{2})}^2=32x-65+1=32x-64$
$\Rightarrow 4{(y+\frac{1}{2})}^2=32(x-2)$
$\Rightarrow {(y+\frac{1}{2})}^2=8(x-2)$
Let $y+\frac{1}{2}=Y$ and $x-2=X$
$\therefore Y^2=8X$
On comparing with $y^2=4ax$ we get $a=2$
According to the question, $X>2a$
$\Rightarrow x-2>4$
or $x>6$
$\therefore x=7,8$
(C)$\because 4y^2-16x-4y+41=0$
$\Rightarrow 4(y^2-y+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2)-16x+41=0$
$\Rightarrow 4{(y-\frac{1}{2})}^2=16x-41+1=16x-40$
$\Rightarrow 4{(y-\frac{1}{2})}^2=8(2x-5)$
$\Rightarrow {(y-\frac{1}{2})}^2=2(x-2)=4(x-\frac{5}{2})$
Let $y-\frac{1}{2}=Y$ and $x-\frac{5}{2}=X$
$\therefore Y^2=4X$
On comparing with $y^2=4ax$ we get $a=1$
According to the question, $X>2a$
$\Rightarrow x-\frac{5}{2}>2$
or $x>\frac{9}{2}=4.5$(approx.)
$\therefore x=5,6,7,8$