Question

The locus of the point $(h,k)$, if the point $(\sqrt{3h},\sqrt{3k+2})$ lies on the line $x-y-1=0$, is a ?

• A. straight line
• B. circle
• C. parabola
• D. none of these

Answer 1

The correct option is C parabola

$(\sqrt{3h},\sqrt{3k+1})$ lines on the line $x-y-1=0$
$\Rightarrow (\sqrt{3h}{)}^2=(\sqrt{3k+2}+1{)}^2$
$\Rightarrow 3h=3k+2+1+2\sqrt{3k+2}$
$\Rightarrow 3^2(h-k-1{)}^2=2^2(\sqrt{3k+2}{)}^2$
$\Rightarrow 9(h^2+k^2+1-2hk-2h+2k)=4(3k+2)$
$\Rightarrow 9(x^2+y^2)-18xy-18x+6y+1=0$
Now $h^2=ab$ and $\Delta \ne 0$
Therefore, locus is a parabola.