Q.174. Curve and its tangent, Rectangular hyperbola

Question 174.
For the curve H with parametric equations x = t, y = 1/t, let S be the point (sqrt 2,sqrt 2). Let N be the point on the tangent to H at P such that SN is perpendicular to PN.
(a) Show that the coordinates of N satisfy the equations (t^2) y + x = 2t and y - (t^2)x = sqrt2 (1 - t^2).
(b) If you square and add the equations in part (a), show that you obtain x^2 + y^2 = 2. Interpret this result geometrically.

Answer 174.
(a)
x = t and y = 1/t
=> dx/dt = 1 and dy/dt = - 1/t^2
=> slope of the tangent, m = dy/dx = (dy/dt) / (dx/dt) = - 1/t^2
=> eqn. of tangent at P is
(y - 1/t) = - 1/t^2 (x - t)
=> x + t^2y = 2t
As N (x, y) belongs to this tangent, its coordinates are given by the equation of the tangent, i.e.,
(t^2) y + x = 2t ... ( 1 )
If SN is perpendicular to PN, its slope = - 1/m = t^2
=> (y - √2) / (x - √2) = t^2
=> y - (t^2) x = √2 (1 - t^2) ... ( 2 )

(b)
Squarring and adding ( 1 ) and ( 2 ),
[(t^2) y + x]^2 + [y - (t^2) x]^2 = 4t^2 + 2(1 - t^2)^2
=> (1 + t^4) y^2 + (1 + t^4) x^2 = 2 [(1 - t^2)^2 + 2t^2]
=> (1 + t^4) (x^2 + y^2) = 2 (1 + t^4)
=> x^2 + y^2 = 2.

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