If the parabolas y^2 = 4b(x - c) and y^2 = 8a | vedclass

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(B) The equation of the normal to the parabola $y^2 = 4b(x - c)$ with slope $m$ is $y = m(x - c) - 2bm - bm^3$.The equation of the normal to the parab

Vedclass · Accepted Answer

$(1, 1, 3)$

Vedclass · Answer

(B) The equation of the normal to the parabola $y^2 = 4b(x - c)$ with slope $m$ is $y = m(x - c) - 2bm - bm^3$.The equation of the normal to the parabola $y^2 = 8ax$ with slope $m$ is $y = mx - 4am - 2am^3$.For these two parabolas to have a common normal with slope $m 
eq 0$,the equations must represent the same line. Comparing the coefficients:$m(x - c) - 2bm - bm^3 = mx - 4am - 2am^3$$-mc - 2bm - bm^3 = -4am - 2am^3$$mc = 4am - 2bm + 2am^3 - bm^3$$mc = m(4a - 2b) + m^3(2a - b)$Dividing by $m$ (assuming $m 
eq 0$):$c = 4a - 2b + m^2(2a - b)$$m^2 = \frac{c - 4a + 2b}{2a - b} = \frac{c - 2(2a - b)}{2a - b} = \frac{c}{2a - b} - 2$.For a common normal to exist,we require $m^2 > 0$,so $\frac{c}{2a - b} > 2$.Testing option $(B)$ $(a=1, b=1, c=3)$:$m^2 = \frac{3}{2(1) - 1} - 2 = \frac{3}{1} - 2 = 1 > 0$.Thus,$(1, 1, 3)$ is a valid choice.

If the parabolas $y^2 = 4b(x - c)$ and $y^2 = 8ax$ have a common normal,then which one of the following is a valid choice for the ordered triad $(a, b, c)$?

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