If the two lines given by $a x^2+2 \mathrm{hxy}+\mathrm{by}^2=0$ make inclinations $\alpha$ and $\beta$, then $ an (\alpha+\beta)=$

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If the two lines given by $a x^2+2 \mathrm{hxy}+\mathrm{by}^2=0$ make inclinations $\alpha$ and $\beta$, then $	an (\alpha+\beta)=$

Aicharya · Accepted Answer

Lines given by ax $2+2 h x y+b^2=0$ make inclinations $\alpha$ and $\beta$. $	herefore 	an \alpha+	an \beta=\frac{-2 \mathrm{~h}}{\mathrm{~b}} 	ext { and } 	an \alpha 	an \beta=\frac{\mathrm{a}}{\mathrm{b}}$ Now $	an (\alpha+\beta)=\frac{	an \alpha+	an \beta}{1-	an \alpha 	an \beta}=\frac{\left(-\frac{2 h}{b}ight)}{1-\left(\frac{a}{b}ight)}=\frac{-2 h}{b} 	imes \frac{b}{(b-a)}$ $	an (\alpha+\beta)=\frac{-2 h}{b-a}=\frac{2 h}{a-b}$

If the two lines given by $a x^2+2 \mathrm{hxy}+\mathrm{by}^2=0$ make inclinations $\alpha$ and $\beta$,…

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