Question

If $x_1$ is the abscissa of the point on the axis of the parabola $3y^2+4y-6x+8=0$ from whlch $3$ distinct normals can be drawn, then

• A. $x_1<\frac{19}{9}$
• B. $x_1>\frac{19}{9}$
• C. $x_1=\frac{19}{9}$
• D. None of these

Answer 1

The correct option is B $x_1>\frac{19}{9}$

$3y^2+4y-6x+8=0$
$\Rightarrow 3(y^2+\frac{4}{3}y).=6x-8$
$\Rightarrow (y+\frac{2}{3}{)}^2=2(x-\frac{4}{3})+\frac{4}{9}=2(x-\frac{10}{9})$
$Y^2=4aX,$ where $Y=y+\frac{2}{3},X=x-\frac{10}{9},a=\frac{1}{2}$
Let the normal through the point $(x_1,\frac{-2}{3})$ on the axis (the axis is $Y=0$ i.e., $y=-\frac{2}{3}$) meet the parabola at $(a{t}^2,2at).$
$\therefore$ its equation is $Y+tX=2at+a{t}^3.$
This passes through $(X_1,0)$ where $X_1=x_1-\frac{10}{9};$ corresponding $Y=-\frac{2}{3}+\frac{2}{3}=0$
$\therefore t(x_1-\frac{10}{9})=2at+a{t}^3\Rightarrow t=0or$
$\Rightarrow a{t}^2=x_1-\frac{10}{9}-2a$
$\Rightarrow x_1-\frac{10}{9}-2\times \frac{1}{2}>0$
$\Rightarrow x_1>\frac{19}{9}$