Answer: [tex]\bold{x^2-\dfrac{181}{13}xy+12y^2=0}[/tex]
Step-by-step explanation:
[tex]\tan \theta =\dfrac{y}{x}\quad \longrightarrow \theta=\tan ^{-1}\bigg(\dfrac{y}{x}\bigg)[/tex]
[tex]\text{Given:}\quad \theta =\tan^{-1}\bigg(\dfrac{1}{13}\bigg)\quad \longrightarrow x=13,y=1[/tex]
Insert x = 13 and y = 1:
x² + 2hxy + 12y² = 0
(13)² + 2(13)(1) + 12(1)² = 0
169 + 26h + 12 = 0
26h + 181 = 0
26h = -181
[tex]h=-\dfrac{181}{26}[/tex]
[tex]\text{Insert:}\ h=-\dfrac{181}{26}[/tex]
[tex]x^2 + 2\bigg(-\dfrac{181}{26}\bigg) xy + 12y^2 = 0[/tex]
[tex]\bold{x^2 -\bigg(\dfrac{181}{13}\bigg) xy + 12y^2 = 0}[/tex]
