maths m2 mock question

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(a) OP^2 = a^2 + (2a^2 - 2)^2

= 4a^4 - 7a^2 + 4

dOP/da = 16a^3 - 14a = 0 => a = 0 or a = √14/4 or -√14/4

The coordinates of P are (√14/4,-1/4)

(b) dy/dx = 4x

At P, dy/dx = √14 and the slope of OP is -1/√14

So, the slope of tangent at P * slope of OP is -1. This means that OP is the normal to the curve C at P

(c) Let |MQ| is the shortest distance from the point Q(5,-1) to the curve C

Let M(p,q)

Slope of tangent at M = 4p

Slope of MQ = (q + 1)/(p - 5)

As 4p * (q + 1)/(p - 5) = -1

4p(q + 1) = 5 - p

4pq = 5 - 5p

But q = 2p^2 - 2

Thus 4p(2p^2 - 2) = 5 - 5p

8p^3 = 5 + 3p

8p^3 - 3p - 5 = 0

(p - 1)(8p^2 - 8p + 5) = 0

p = 1 is the only real root

So, M(1,0) and MQ = √17

c) Let D = distance (PQ)

D = √ { (a-5)^2 + [2(a^2) -2 - (-1)]^2}
= (4a^4 - 3a^2 -10a +26)^(1/2)

dD/da = (1/2)* (4a^4 - 3a^2 -10a +26)^(-1/2) * (16a^3 - 6a - 10)
= (8a^3 - 3a - 5) * (a^4 - 3a^2 -10a +26)^(-1/2)

when dD/da = 0,
(8a^3 - 3a - 5) = 0
(a-1)(8a^2 + 8a +5) = 0
a = 1 is the only real root

so the required distance
= D(1)
= (4*1^4 - 3*1^2 -10*1 +26)^(1/2)
= √17 #