Help - Tangents and Normals: Parametric Approach (1 Viewer)

[candychu]

can someone pls help me with these questions:

3 unit -
Tangents and Normals: Parametric Approach

If y = 1-2x is a tangent to x^2 = 4ay, find a and the point of contact.

This is wat i did:
solve simulataneously to get:
x^2 = 4a(1-2x)
x^2 + 8ax - 4a = 0 -- (1)

let discriminant = 0
64a^2 = -16a
a= -1/4

sub a into (1)
x^2 - 2x + 1 = 0
(x-1)^2 = 0
so x = 1 and y = -1

but the answer is (-1, 1) as point of contact

so i tried subbing a= -1/4 into x^2 = 4ay
and i got y = x^2
solve simultaneously with y = 1-2x

then i get x^2 + 2x - 1 = 0
and u cant solve that ><"
unless u complete the square and u get
(x + 1)^2 = 3
so x can't equal -1 *shrug*


i dont know wat ive done wrong? can anyone help me?<!--colorc--><!--/colorc-->


<!--coloro:#FF0000--><!--/coloro-->
another question ><"
a)Show that L:ax+by=1 is tangent to P: x^2=12y when 3a^2 + b=0
this question i cud do but there r 2 more parts

b)Hence find the tangents at P wiht y-int : -27
c) Show that if L passes through U(4,1) then 4a + b = 1. Hence find the tangets to P though U.<!--colorc--><!--/colorc-->


thanx in advance~

 

[z600]

I did the first question and had the same outcome as you did but the answer co-ordinates makes sense. The straight line has a negative gradient and thus it must touches the parabola on the negative side.

I will come back and do 2
a) ax+b(x^2/12)=1
bx^2+12ax-12=0

Take delta (discriminant) and it equal to 0
then you get 3a^2+b=0

b) using family of lines

(y+27)=mx
y=mx-27

sub into the equation and you get
x^2-12m-324=0
take discriminant and let it equal to 0
144m^2+1296=0
m=-3 or +3
Sub m back to family of lines

y+27=-3x or y+27=3x

am i right here?

c) with that i am sure you can work c out

All these questions out of cambridge?

 

[candychu]

hey! yes theyre out of cambridge
thanx for helping me ^.^
yup the answers to the 2nd question is right

lol so the first question is a bit of an odd one eh? *confused*

 

[SoulSearcher]

Actually, by having a = -1/4, your answer makes sense. If you draw that particular curve, the parabola would be concave down, meaning that y = 1 - 2x would only be a tangent to the curve on the positive side of the x-axis, which is in agreement with the value of a = -1/4 and the resulting curve.