If a, b and c form a geometric progression wi | vedclass

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(C) Since $a, b, c$ are in geometric progression,we have $b = ar$ and $c = ar^2$. Substituting these into the line equation $ax + by + c = 0$,we get $

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$\frac{r}{2}$

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(C) Since $a, b, c$ are in geometric progression,we have $b = ar$ and $c = ar^2$. Substituting these into the line equation $ax + by + c = 0$,we get $ax + ary + ar^2 = 0$. Dividing by $a$ (assuming $a 
eq 0$),we get $x + ry + r^2 = 0$,or $x = -ry - r^2$. Substituting this into the curve equation $x + 2y^2 = 0$,we get $(-ry - r^2) + 2y^2 = 0$. Rearranging gives the quadratic equation $2y^2 - ry - r^2 = 0$. The sum of the ordinates ($y$-coordinates) is the sum of the roots of this quadratic equation,which is given by $-\frac{	ext{coefficient of } y}{	ext{coefficient of } y^2} = -(\frac{-r}{2}) = \frac{r}{2}$.

If $a, b$ and $c$ form a geometric progression with common ratio $r$,then the sum of the ordinates of the points of intersection of the line $ax + by + c = 0$ and the curve $x + 2y^2 = 0$ is

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