Unknown

Solutions to Problems; Chapter (^1 257)
where tan+ = (a - b)/2h. The maximum (minimum) occurs when
sin(20 + 4) is equal to 1 (-1, respectively).
8.20. f(x, y) h as 1’ mear factors if and only if f(x, y) = 0 is solvable in x
as a linear function of y. The discriminant of this quadratic equation is
4y2 - Shy + 4a2 + 4c, which is a square of a linear function in y exactly
when b2 = a2 + c.
8.21. (a) Substituting y = mx + c into b2x2 - a2y2 = a2b2 yields the
quadratic equation
(a2m2 - b2)x2 + 2a2cmx + a2(b2 + c”) = 0
for the abscissae of the intersection point. The discriminant is
4u2b2(b2 + c2 - a2m2) and this vanishes if and only if the line is tangent to
the curve.
(b) The midpoint between the points of intersection of the line y = mx+c
and the circle x2 + y2 = r2 is
(
-- -cm c
1+m2’l+m2 > ’
Let y = mx + c be tangent to b2x2 - a2y2 = a2b2, so that a2m2 = b2 + c2
and let (x, y) be on the locus. Then
m = -z/y c = (x2 + y2)/y.
Hence a2x2 = b2y2 + (x2 + y2)2. Note that this is independent of r and
contains the points (0,O) and (a, 0). However, to take account of the fact
that the points on the locus lie within the circle, we require that x2 + y2 5
r2.
8.22. at2 + bt + c has zeros r + is, -r + is (r # 0) if and only if at2 - ibt - c
has zeros s + ir, s - ir, which occurs if and only if ib/a, c/a are real
and b2 - 4ac 1 0. Also, at2 + bt + c has zeros iu and iv if and only if
at2 - ibt - c has real zeros, which occurs if and only if ib/u and c/u are real
and b2 - 4ac < 0. Hence, the necessary and sufficient conditions are that
ib/a and c/u are real.
8.23. If one diameter is y = mx, its conjugate is ma2y+ b2x = 0. The line
y = mx intersects the ellipse in points (~1, v) for which (b2+a2m2)u2 = a2b2;
the line ma2y+b2x = 0 in points (z, w) for which (m2a2+b2)z2 = m2a4. We
require that u2 + v2 = z2 + w2, which leads to m2a2( b2 - a2) = b2(b2 - u2).
The case a = b is that of a circle and all diameters are of equal length.
When a # b, we must have m = b/a, so that the equations of the conjugate
diameters are bx + ay = 0 and bx - ay = 0.
8.24. Invoking the given inequality for x = 0, f, 1 yields -1 5 c 5 1,
-4 5 a + 2b + 4c 5 4, -1 5 a + b + c 5 1, respectively. Eliminating b and