Show that the locus of the middle points of portions of the tangents to the hyperbola x 2 a 2 - y 2 b 2 = 1 intercepted between the axes is 4 x 2 y 2 = a 2 y 2 - b 2 x 2 .

x 2 a 2 - y 2 b 2 = 1 P = ( a sec θ , b tan θ ) Equation of tangent at P , x sec θ a - y tan θ b = 1 , x - intercept. x = a sec θ , y = intercept. y = b tan θ Points on axes are Q = a sec θ 0 & R = 0 , b tan θ Now mid -point M = h , k = a 2 sec θ , b 2 tan θ ⇒ sec θ = a 2 h & tan θ = b 2 k Now sec 2 θ - tan 2 θ = 1 ⇒ a 2 4 h 2 - b 2 4 k 2 = 1 ⇒ a 2 y 2 - b 2 x 2 = 4 x 2 y 2 . Hence, proved.

Show that the locus of the foot of perpendicular drawn from the centre of the hyperbola x 2 a 2 - y 2 b 2 = 1 , on any tangent to it is x 2 + y 2 2 = a 2 x 2 - b 2 y 2 .

Given the equation of a hyperbola x 2 a 2 - y 2 b 2 = 1 , whose centre will be ( 0 , 0 ) . Let a point on the tangent of the hyperbola which is foot of perpendicular drawn from the origin is ( h , k ) . As we know the equation of the tangent on hyperbola is written as y = m x ± a 2 m 2 - b 2 , where m , is the slope of the tangent to the hyperbola. As we know the equation of line which is perpendicular to a given line a x + b y + c = 0 , is b x - a y = k . So the equation of a line which is normal to the hyperbola is written as x + m y = λ . As this normal line passes from the origin, so the origin satisfies the equation of the normal. ⇒ 0 + 0 = λ . ⇒ λ = 0 . Again this line also passes from the point ( h , k ) . ⇒ h + m k = 0 . ⇒ m = - h k . As the equation of tangent to hyperbola passes from the point ( h , k ) . ⇒ k = m h ± a 2 m 2 - b 2 . ⇒ k 2 + h 2 k = a 2 h 2 - k 2 b 2 k 2 . Squaring both side we get. ⇒ k 2 + h 2 k 2 = a 2 h 2 - k 2 b 2 k 2 2 . ⇒ k 2 + h 2 2 = a 2 h 2 - k 2 b 2 . So the locus of the foot of perpendicular from the centre of the hyperbola on the tangent to the hyperbola is y 2 + x 2 2 = a 2 x 2 - y 2 b 2