geometry help 2?

Question

1.Find the slope of the tangent of the hyperbola

x²/16-y²/4=1

at the point R(5,-3/2)


A.5/6
B.-(5/6)
C.6/5
D.-(6/5)


2.Write in polar coordinates the equation y=3x+5

A.r=3 cos 0+5
B.r=3sin 0 +5 cos 0
C.r=5/ sin 0-3 cos 0
D.r= 3/sin 0-cos 0

Answer 1

Part 1
The tangent at a point (a, b) to the hyperbola

x²/16 - y²/4 = 1

has the equation

ax/16 - by/4 = 1

(similarly for other 2 deg curves). You can see why this is a straight line passing thru (a, b). It's less obvious that this is the tangent line (to show this, work out the gradient vectors of x²/16 - y²/4 and of ax/16 - by/4 at (a, b); they are the same up to scale, which means that the hyperbola and the straight line are perpendicular to the same vector, and therefore are tangent to each other).

At the point a = 5 and b = -3/2, the line's slope is a/4b = -5/6.


In Part 2, the "0" must mean the angle theta.
In polar coordinates,

x = r cos(theta)
y = r sin(theta)

So the equation y = 3x + 5 becomes

r sin(theta) = 3r cos(theta) + 5

or

r = 5/[sin(theta) - 3cos(theta)]

(which is C).

You can also use trig identities to rewrite this in the form r = D/cos(theta - a constant), which is the std form of the polar equation of a straight line (see, for example, http://www.xtec.es/~rgonzal1/treatise.pdf, page 41).

Answer 2

1.Find the slope of the tangent of the hyperbola x²/16-y²/4=1 at the point R(5,-3/2)

x² - 4y² = 16
-4y² = 16 - x²
y² = x²/4 - 4
y = +/- sqrt [x²/4 - 4]
y' = +/- 1/2 [ (x²/4 - 4)^-(1/2) ] (x²/4 - 4)'
y' = +/- 1/2 [ (x²/4 - 4)^-(1/2) ] x/2
y' = +/- x/4 [ (x²/4 - 4)^-(1/2) ]
y' = +/- 5/4 [ (5²/4 - 4)^-(1/2) ]
y' = +/- 5/4 [ (25/4 - 4)^-(1/2) ]
y' = +/- 5/4 [ (9/4)^-(1/2) ]
y' = +/- 5/4 [ (4/9) ^ (1/2) ]
y' = +/- 5/4 [ 2/3 ]
y' = +/- 10/12
y' = +/- 5/6
but since the hyperbolic coordinates are under the axis, that rule out any positive slopes. Therefore:
b) is your answer [ -5/6]

2.Write in polar coordinates the equation y=3x+5
(I think your missing "when x = 0" or "when y = 0" ) because all the answers contain 0. So be it!

If x=0, y=3(0) + 5 then (0,5) so it would be 3sin0 + 5 and the 3 is arbitrary because sin0 = 0 so the 3 could be 1,000,000,000 and it wouldn't matter.

If y=0, 0=3x + 5 then (-5/3,0) so it would be "cos0 - (8/3)" or "sin0 - (5/3)"