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 Question from Gareth, a student: 2. Consider the rectangular hyperbola xy = c^2 with parameterization (x, y) = (ct, c/t), and t 6= 0. i. Derive (i.e. do not just quote) the equations of the tangent and normal at the point P with (parametric) coordinates (cp, c/p). ii. Consider the points P : (cp, c/p) and Q : (cq, c/q) on the hyperbola. Find the equation of the straight line that joins the points P and Q. iii. Consider a point R : (cr, c/r) also on the hyperbola, and suppose that in the triangle PRQ, \PRQ = 90◦. Prove that the normal (to the hyperbola) at R is parallel to the line PQ.

Gareth,

1. The equation of the hyperbola is y = c2/x = c2 x-1 and hence dy/dx = - c2 x-2 = - c2/x2. Thus at P where x = cp, the slope of the tangent line is - c2/(cp)2 = - 1/p2. The tangent line at P is hence the line through (cp, c/p) and slope - 1/p2. The normal line at P is then the line trough (cp, c/p) with slope p2.

2. Write the equation of the line through P (cp, c/p) and Q (cq, c/q).

3. In ii. you found the slope of the line joining P and Q so you need to show that the normal at R has the same slope. From i. you know the slope of the normal at R but this is in terms of r where the slope of the line joining P and Q is in terms of p and q. Write m1, the slope of the line joining P and R and m2, the slope of the line joining Q and R. Since these lines are perpendicular, the slopes must be negative reciprocals. This will give you a relationship between m1 and m2 which will allow you to write the slope of the normal at R in terms of p and q.

I hope this helps,
Harley

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