Answer to Question #157493 in Calculus for Carty

Question

Given that the equation normal to the rectangular hyperbola xy = c^2 at the point P(ct, c/t) is t^3*x - ty = c(t^4 - 1)
The normal at P on the hyperbola meets the x - axis at Q and the tangent at P meets the y - axis at R. Show that the locus of the midpoint of QR, as P varies, is
2c^2*xy + y^4 = c^4

Answer 1

The equation of normal at the point Q:

$t^3x=c(t^4-1)$

Then point Q: $(\frac{c(t^4-1)}{t^3},0)$

The equation of the tangent at the point P:

$y-y_P=f'(P)(x-x_P)$

$y'(x)=-\frac{c^2}{x^2}$

$f'(P)=-\frac{c^2}{c^2t^2}=-\frac{1}{t^2}$

Then:

$y-c/t=-\frac{1}{t^2}(x-ct)$

At the point R:

$y-c/t=c/t$

Point R: $(0, 2c/t)$

Midpoint of QR: $(\frac{c(t^4-1)}{2t^3}, \frac{c}{t})$

Then, for midpoint of QR:

$2c^2xy + y^4=\frac{c^4(t^4-1)}{t^4}+\frac{c^4}{t^4}=\frac{c^4t^4}{t^4}=c^4$