If the two lines given by a x^2+2 hxy+by^2=0 make inclinations α and β , then tan (α+β)=

Correct answer: (D) (2 h)/(a-b). Lines given by ax 2+2 h x y+b^2=0 make inclinations α and β . ∴ tan α+tan β=(-2 ~h)/(~b) and tan α tan β=a/b Now tan (α+β)=(tan α+tan β)/(1-tan α tan β)=((-(2 h)/b ))/(1- (a/b ))=(-2 h)/b × b/((b-a)) tan (α+β)=(-2 h)/(b-a)=(2 h)/(a-b)

Q85 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics

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

A
$\frac{\mathrm{h}}{\mathrm{a}+\mathrm{b}}$
B
$\frac{2 h}{a+b}$
C
$\frac{h}{a-b}$
Correct answer
D
$\frac{2 h}{a-b}$

Solution

Lines given by ax $2+2 h x y+b^2=0$ make inclinations $\alpha$ and $\beta$. $$ \therefore \tan \alpha+\tan \beta=\frac{-2 \mathrm{~h}}{\mathrm{~b}} \text { and } \tan \alpha \tan \beta=\frac{\mathrm{a}}{\mathrm{b}} $$ Now $\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{\left(-\frac{2 h}{b}\right)}{1-\left(\frac{a}{b}\right)}=\frac{-2 h}{b} \times \frac{b}{(b-a)}$ $$ \tan (\alpha+\beta)=\frac{-2 h}{b-a}=\frac{2 h}{a-b} $$

Hence, the correct answer is option: 'D'.

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

Solution

Relevant Questions