Question

In a plane rectangular coordinate system, there are three points $A(0,\frac{4}{3}),B(–1,0)$ and $C(1,0).$ The distance from point $P$ to line $BC$ is the geometric mean of the distances from this point to lines $AB$ and $AC.$ Then the locus of point $P$ is

• A. $2x^2+2y^2+3y-2=0$
• B. $2x^2-2y^2-3y+2=0$
• C. $8x^2-17y^2+12y-8=0$
• D. $8x^2+17y^2-12y-8=0$

Answer 1

Answer: (A), (C)

$8x^2-17y^2+12y-8=0$

Equation of line $AB:$
$y=\frac{4}{3}(x+1)$
$\Rightarrow 4x-3y+4=0$
Equation of line $AC:$
$y=-\frac{4}{3}(x-1)$
$\Rightarrow 4x+3y-4=0$
Equation of line $BC:$
$y=0$

Let point $P=(\alpha ,\beta ).$ Then,
Distance of point $P$ from $AB:$
$d_1=\frac{1}{5}|4\alpha -3\beta +4|$
Distance of point $P$ from $AC:$
$d_2=\frac{1}{5}|4\alpha +3\beta -4|$
Distance of point $P$ from $BC:$
$d_3=|\beta |$

According to question,
$d_1{d}_2=d_3^2$
$\Rightarrow |16{\alpha }^2-(3\beta -4{)}^2|=25{\beta }^2$
$\Rightarrow 16{\alpha }^2-(3\beta -4{)}^2=\pm 25{\beta }^2$

Taking positive sign, we get
$16{\alpha }^2-9{\beta }^2-16+24\beta =25{\beta }^2$
$\Rightarrow 8{\alpha }^2-17{\beta }^2+12\beta -16=0$
$\therefore$ Locus of point $P$ is
$8x^2-17y^2+12y-8=0$

Taking negative sign, we get
$2{\alpha }^2+2{\beta }^2+3\beta -2=0$
$\therefore$ Locus of point $P$ is
$2x^2+2y^2+3y-2=0$