The correct options are
A √2(x−3y)=4
B √2(x−3y)=−4
Differentiating the equation of thehyperbola w.r.t x, we get:
6x−2ydydx=0or dydx=3xyLet the point on the curve where the normal has slope 1/3 be P(x1,y1).Hence, −dxdy=−y13x1=13 or y1=−x1Also, P lies on the hyperbola∴3x12−y12=1⇒3x12−x12=1⇒x1=±1√2Also, y1=∓1√2Hence, the points on the curve are(±1/√2, ∓1/√2)Hence, the equations of the normals are:y−1√2=13(x+1√2)and y+1√2=13(x−1√2)⇒√2(x−3y)=4 and √2(x−3y)=−4