Question

The points on the axis of the parabola $3y^2+4y-6x+8=0$ from where 3 distinct normals can be drawn are given by

• A. $(a,\frac{2}{3}),a>\frac{19}{9}$
• B. $(a,-\frac{2}{3});a>\frac{19}{9}$
• C. $(a,\frac{2}{3});a>\frac{7}{9}$
• D. $(a,-\frac{2}{3});a>\frac{20}{9}$

Answer 1

The correct option is B $(a,-\frac{2}{3});a>\frac{19}{9}$

The points on the axis of parabola $3y^2+4y-6x+8=0$ from where $3$ distinct normals can be drawn are given by

$3(y^2+\frac{4}{3}y)=6x-83(y+\frac{2}{3})=6x-8+\frac{4}{3}{(y+\frac{2}{3})}^2=\frac{6}{3}(x-\frac{10}{9})=2(x-\frac{10}{9}).............\left(1\right)$

Axis of parabola is $y=-\frac{2}{3}$

Any point on axis is $(h,-\frac{2}{3})$

Written (1) in new form

$y^2=4\left(\frac{1}{2}\right)x$

Any point on new axis is $(H,0)$

If it has $3$ normals

Then, $H>2a$ (property)

$H>2\left(\frac{1}{2}\right)=1h-\frac{10}{9}>1h>\frac{10}{9}$

So Coordinate is $(a,-\frac{2}{3});a>\frac{19}{9}$