Question

Consider three lines y axis, y = $2$ and x + my = $1$ where (l, m) lies on $y^2=4x$. Locus of circum centre of triangle formed by given three lines is a parabola whose vertex is

• A. $(-2,\frac{3}{2})$
• B. $(2,\frac{-3}{2})$
• C. $(-2,\frac{-3}{2})$
• D. $(2,\frac{-5}{2})$

Answer 1

The correct option is A $(-2,\frac{3}{2})$

Given lines are y -axis $(x=0),y=2$ and

$x+my=l$

The rough diagram will line.

$\Rightarrow$ point A is

$y=2\Rightarrow x=l-2m\Rightarrow A(l-2n,2)$

point C is $x=0\Rightarrow y=\frac{l}{m}\Rightarrow c(0,\frac{l}{m})$

$△ABC$ is the right angel triangle

$\Rightarrow$ circum centre of $△ABC$ is mid point of AC

[mid point of hypotenuse is the circum centre]

$\Rightarrow$ Let mid point is P(h, k)

$\Rightarrow h=\frac{0+l-2m}{2}$ & $k=\frac{2+\frac{l}{m}}{2}$

$\Rightarrow 2h=l-2m$ & $2k=2+\frac{l}{m}...\left(1\right)$

(l,m) lies on parabola $y^2=4x$

$\Rightarrow \left(lm\right)\equiv (t^2,2t)$ point in parameter form.

$\Rightarrow$ (1)statement becomes.

$2h=t^2-4t$ & $2k=2+\frac{t^2}{2t}=2+\frac{t}{2}$

from $2k=2+\frac{t}{2}\Rightarrow 4k=4+t\Rightarrow t=4k-4$

putting in $2h=t^2-4t$

$\Rightarrow 2h+4=t^2-4t+4$

$\Rightarrow 2h+4=(t-2{)}^2$

$(2h+4)=(4k-6{)}^2$

$\Rightarrow$ Equation of parabola is

$2x+4=(4y-6{)}^2$

$\Rightarrow 4^2(y-\frac{3}{2}{)}^2=2(x+2)$

$\Rightarrow {(y-\frac{3}{2})}^2=\frac{1}{8}(x-(-2\left)\right)$

$\Rightarrow$ vertex is $(-2,\frac{3}{2})$