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Standard XII

Mathematics

Condition for Two Lines to Be Perpendicular

If a,b,c are ...

Question

If a,b,c are in GP and the line $a x + b y + c = 0$ and curve $x + 2 y^{2} = 0$ intersects at two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ and $2 (y_{1}, y_{2}) == \frac{b^{α}}{a^{β}}$ , then $2 α + 3 β$ equals

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Solution

$a, b, c$ are in GP let $b = a r a n d c = a r^{2}$
So line is $a x + a r y + a r^{2} = 0$ now curve is $x + 2 y^{2} = 0$
Eliminating a we get $2 y^{2} - r y - r^{2}$
So $y_{1} + y_{2} = \frac{r}{2}$
So $2 (y_{1} + y_{2}) = r = \frac{b}{a}$

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If a,b,c are in GP and the line ax+by+c=0 and curve x+2y2=0 intersects at two points (x1,y1) and (x2,y2) and 2(y1,y2)==bαaβ, then 2α+3β equals

a,b,c are in GP let b=arandc=ar2So line is ax+ary+ar2=0now curve is x+2y2=0 Eliminating a we get 2y2−ry−r2So y1+y2=r2So 2(y1+y2)=r=ba

If a,b,c are in GP and the line $a x + b y + c = 0$ and curve $x + 2 y^{2} = 0$ intersects at two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ and $2 (y_{1}, y_{2}) == \frac{b^{α}}{a^{β}}$ , then $2 α + 3 β$ equals

$a, b, c$ are in GP let $b = a r a n d c = a r^{2}$
So line is $a x + a r y + a r^{2} = 0$ now curve is $x + 2 y^{2} = 0$
Eliminating a we get $2 y^{2} - r y - r^{2}$
So $y_{1} + y_{2} = \frac{r}{2}$
So $2 (y_{1} + y_{2}) = r = \frac{b}{a}$