Hyperbola
Transverse axis from asymptotes
MJAT_TS5_P1
Grade 12
Question:
The equation of the transverse axis of a hyperbola (passing through origin) having asymptotes $3x-4y-1=0$ and $4x-3y-6=0$ is $ax+by-c=0$, where $a,b,c\in\mathbb{N}$ and $\gcd(a,b,c)=1$. Then $(a+b+c)=$
Step-by-Step Solution
Key Concept: Transverse axis bisects the angle between asymptotes (the acute angle bisector for the transverse axis). The two asymptotes are $L_1: 3x-4y-1=0$ and $L_2: 4x-3y-6=0$. The angle bisectors are $\frac{3x-4y-1}{5}=\pm\frac{4x-3y-6}{5}$.
$a+b+c=\mathbf{7}$.
Correct Answer: 7