12.43
Show that the ellipse 4x
2 + 9y
2 = 45 and the hyperbola x
2 — 4y
2 = 5 are orthogonal.
I To find the intersection points, multiply the equation of the hyperbola by 4 and subtract the result from the
equation of the ellipse, obtaining 25y
2 = 25, y
2 = l, y = ±l, x = ±3. Differentiate both sides of the
equation of the ellipse: 8* + I8yy' = 0, y' — — (4x/9y), which is the slope of the tangent line. Differentiate
both sides of the equation of the hyperbola: 2x — 8yy' = 0, y' = x/4y, which is the slope of the tangent line.
Hence, the product of the slopes of the tangent lines is -(4x/9y)- (x/4y) — — (x
2 /9y
2 ). Since x
2 = 9 and
y
2 = 1 at the intersection points, the product of the slopes is —1, and, therefore, the tangent lines are
perpendicular.
12.44
Find the slope of the tangent line to the curve x
2 + 2xy — 3y
2 = 9 at the point (3,2).
I 2x + 2(xy'+ y)-6yy'=0. Replace x by 3 and y by 2, obtaining 6 + 2(3y' + 2) - I2y' =0, 10-6y'=0,
y' = 3 . Thus, the slope is § .
12.45
Show that the parabolas y
2 = 4x + 4 and y
2 = 4 — 4x intersect at right angles.
I To find the intersection points, set 4x + 4 = 4-4x. Then x = 0, y
2 =4, y = ±2. For the first
parabola, 2yy'=4, y' = 2/y. For the second parabola, 2yy' = -4, y'=—2/y. Hence, the product of
the slopes of the tangent lines is (2/y)(-2/y) = -4/y
2 . At the points of intersection, y
2 = 4. Hence,,the
product of the slopes is —1, and, therefore, the tangent lines are perpendicular.
12.46
Show that the circles x
2 + y
2 - I2x - 6y + 25 = 0 and x
2 + y' + 2x + y - 10 = 0 are tangent to each other
at the point (2,1).
I For the first circle, 2x + 2yy' - 12 -6y' = 0, and, therefore, at (2,1), 4 + 2y' - 12 - 6y' =0, y'=-2.
For the second circle, 2x + 2yy' + 2 + y' = 0, and, therefore, at (2,1), 4 + 2/ + 2 + y' = 0, >•' = -2.
Since the tangent lines to the two circles at the point (2,1) have the same slope, they are identical, and, therefore,
the circles are tangent at that point.
12.47
If the curve sin y = x* - x
5
passes through the point (1,0), find y' and y" at the point (1, 0).
I (cos _>'))'' = 3x
2 - 5x
4 . At (1,0), y'=3 —5 = —2. Differentiating again, (cos y)y" — (sin y)y'=• 6.v -
20x
3 . So, at (1,0), / = 6-20 =-14.
12.48
If x + y = xy, show that y" = 2y*lx\
I 1 + y' = xy' + y, y'(l — x) = y — 1. Note that, from the original equation, y — l = y/x and .v — 1 =
x/y. Hence, y' = -y
2 /x
2 . From the equation y'(l—x) — y-l, y'(—l) + y"(l — x) = y', y"(\-x) = 2y',
y"(-x/y) = 2(-y
2 /x
2 ), y" = 2y
3 /x*.
80
CHAPTER 12
Show that the ellipse 4x
2 + 9y
2 = 45 and the hyperbola x
2 — 4y
2 = 5 are orthogonal.
I To find the intersection points, multiply the equation of the hyperbola by 4 and subtract the result from the
equation of the ellipse, obtaining 25y
2 = 25, y
2 = l, y = ±l, x = ±3. Differentiate both sides of the
equation of the ellipse: 8* + I8yy' = 0, y' — — (4x/9y), which is the slope of the tangent line. Differentiate
both sides of the equation of the hyperbola: 2x — 8yy' = 0, y' = x/4y, which is the slope of the tangent line.
Hence, the product of the slopes of the tangent lines is -(4x/9y)- (x/4y) — — (x
2 /9y
2 ). Since x
2 = 9 and
y
2 = 1 at the intersection points, the product of the slopes is —1, and, therefore, the tangent lines are
perpendicular.
12.44
Find the slope of the tangent line to the curve x
2 + 2xy — 3y
2 = 9 at the point (3,2).
I 2x + 2(xy'+ y)-6yy'=0. Replace x by 3 and y by 2, obtaining 6 + 2(3y' + 2) - I2y' =0, 10-6y'=0,
y' = 3 . Thus, the slope is § .
12.45
Show that the parabolas y
2 = 4x + 4 and y
2 = 4 — 4x intersect at right angles.
I To find the intersection points, set 4x + 4 = 4-4x. Then x = 0, y
2 =4, y = ±2. For the first
parabola, 2yy'=4, y' = 2/y. For the second parabola, 2yy' = -4, y'=—2/y. Hence, the product of
the slopes of the tangent lines is (2/y)(-2/y) = -4/y
2 . At the points of intersection, y
2 = 4. Hence,,the
product of the slopes is —1, and, therefore, the tangent lines are perpendicular.
12.46
Show that the circles x
2 + y
2 - I2x - 6y + 25 = 0 and x
2 + y' + 2x + y - 10 = 0 are tangent to each other
at the point (2,1).
I For the first circle, 2x + 2yy' - 12 -6y' = 0, and, therefore, at (2,1), 4 + 2y' - 12 - 6y' =0, y'=-2.
For the second circle, 2x + 2yy' + 2 + y' = 0, and, therefore, at (2,1), 4 + 2/ + 2 + y' = 0, >•' = -2.
Since the tangent lines to the two circles at the point (2,1) have the same slope, they are identical, and, therefore,
the circles are tangent at that point.
12.47
If the curve sin y = x* - x
5
passes through the point (1,0), find y' and y" at the point (1, 0).
I (cos _>'))'' = 3x
2 - 5x
4 . At (1,0), y'=3 —5 = —2. Differentiating again, (cos y)y" — (sin y)y'=• 6.v -
20x
3 . So, at (1,0), / = 6-20 =-14.
12.48
If x + y = xy, show that y" = 2y*lx\
I 1 + y' = xy' + y, y'(l — x) = y — 1. Note that, from the original equation, y — l = y/x and .v — 1 =
x/y. Hence, y' = -y
2 /x
2 . From the equation y'(l—x) — y-l, y'(—l) + y"(l — x) = y', y"(\-x) = 2y',
y"(-x/y) = 2(-y
2 /x
2 ), y" = 2y
3 /x*.
80
CHAPTER 12