The point on the axis of the parabola 3y^2+4y | vedclass

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Question

(B) The given equation of the parabola is $3y^2 + 4y - 6x + 8 = 0$. Rearranging the terms: $3(y^2 + \frac{4}{3}y) = 6x - 8$. Completing the square: $3

Vedclass · Accepted Answer

$\left( a, -\frac{2}{3} \right); a > \frac{19}{9}$

Vedclass · Answer

(B) The given equation of the parabola is $3y^2 + 4y - 6x + 8 = 0$. Rearranging the terms: $3(y^2 + \frac{4}{3}y) = 6x - 8$. Completing the square: $3(y^2 + \frac{4}{3}y + \frac{4}{9}) = 6x - 8 + \frac{4}{3}$. $3(y + \frac{2}{3})^2 = 6x - \frac{20}{3}$. $(y + \frac{2}{3})^2 = 2(x - \frac{10}{9})$. This is of the form $Y^2 = 4AX$,where $Y = y + \frac{2}{3}$,$X = x - \frac{10}{9}$,and $4A = 2 \implies A = \frac{1}{2}$. The axis of the parabola is $Y = 0$,which implies $y = -\frac{2}{3}$. Any point on the axis is $(a, -\frac{2}{3})$. For a point $(X_0, 0)$ on the axis of $Y^2 = 4AX$ to have $3$ real normals,we must have $X_0 > 2A$. Substituting the values: $x - \frac{10}{9} > 2(\frac{1}{2}) = 1$. $x > 1 + \frac{10}{9} = \frac{19}{9}$. Thus,the point is $(a, -\frac{2}{3})$ where $a > \frac{19}{9}$.

The point on the axis of the parabola $3y^2+4y-6x+8=0$ from which $3$ real normals can be drawn is given by:

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