If $a, b$ and $c$ form a geometric progression with common ratio $r$, then the sum of the ordinates of the…
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 $a x+b y+c=0$ and the curve $x+2 y^2=0$ is
$-\frac{r^2}{2}$
$-\frac{r}{2}$
$\frac{r}{2}$
$r$
Solution
Since, $a, b$ and $c$ form a geometric progression
$\therefore \quad a=a, b=a r, c=a r^2$
Therefore, given line becomes
$\begin{aligned} & a x+a r y+a r^2 & =0 \\ \Rightarrow & x+r y+r^2 & =0 \\ \Rightarrow & x & =-r y-r^2\end{aligned}$
On putting $x=-r y-r^2$ in given curve $x+2 y^2=0$, we get
$\begin{aligned}-r y-r^2+2 y^2 & =0 \\ \Rightarrow \quad 2 y^2-r y-r^2 & =0\end{aligned}$
$\therefore$ Sum of ordinates $=\frac{r}{2}$