General form of hyperbola is
−=1 ...(i)
Given equation of hyperbola is
2x2−y2=2 ⇒x2−=1 ...(ii)
comparing equation (i) with (ii), we get
a2=1,b2=2 General equation of normal,
+=a2+b2 ...(iii)
Equation of normal at point
A(secθ,2tanθ) +=1+2 ⇒+=3 ⇒+=3 ⇒+ycosθ×cosecθ=3 ⇒+cosecθ=3 ⇒x+ycosecθ=3secθ ...(iv)
Similarly, equation of normal at point
B(secϕ,2tanϕ) x+ycosecϕ=3secϕ ⇒x+ycosec(−θ)=3sec(−θ) ⇒x+ysecθ=3cosecθ ...(v)
subtract equation (v) from (iv), we get
⇒y(cosecθ−secθ)=3(secθ−cosecθ) ⇒y(cosecθ−secθ)=−3(cosecθ−secθ) ⇒y=−3 Now, from equation (ii), we get
x2−=1 x2−=1 [∵y=−3]
x2=1+ x=√ ∴(x,y)=(α,β)=(√,−3) ∴(2β)2=(2×(−3))2=36
