The correct options are
A x−3y=2√2
B x−3y=−2√2
3x2−y2=1
⇒x2(13)−y2=1
a2=(13), b2=1
The equation of normal at any point (1√3secθ,tanθ) on the hyperbola is given by
1√3xsecθ+ytanθ=43
x√3cosθ+ycotθ=43
Slope = −cosθ√3×sinθcosθ⇒−sinθ√3=13
sinθ=−1√3
⇒θ is in 3rd or 4th quadrant.
If θ is in 3rd quadrant ⇒tanθ=1√2, secθ=−√3√2
If θ is in 4th quadrant ⇒tanθ=−1√2, secθ=√3√2
The equations of normals are
√2x3−√2y=43
or −√2x3+√2y=43
⇒x−3y=2√2
or x−3y=−2√2