cant figure this q. out... (need help) (1 Viewer)

[blackfriday]

i did lots of working and still got no where

show that the line x = m(y-1) + 2/m is a tangent to the parabola x^2 = 8(y-1)

much help appreciated

 

[velox]

Differentiate the line and then solve simultaneously with the parabola. You should find a common point.

 

[acmilan]

i did lots of working and still got no where

show that the line x = m(y-1) + 2/m is a tangent to the parabola x^2 = 8(y-1)

much help appreciated

x = m(y-1) + 2/m
y = (x/m) - (2/m^2) + 1....1
x^2 = 8(y-1)....2

sub 1 into 2

x^2 = 8[(x/m) - (2/m^2)]
x^2 = (8x/m) - (16/m^2)
m^2x^2 - 8mx + 16 = 0

Now if the line is a tangent to the parabola then
b^2 - 4ac = 0
LHS
= 64m^2 - 4m^2.16
= 64m^2 - 64m^2
= 0

hence x = m(y-1) + 2/m is a tangent to x^2 - 8(y-1)

 

[blackfriday]

cheers. thanks heaps.

i tried squaring x and subbing in but i got some weird stuff after i worked out the discriminant.

 

[velox]

Hmm i dont get the discriminant bit?

 

[blackfriday]

wrx: i actually couldnt figure out how to diff. it with three variables.

 

[velox]

Why are we using the discriminant?

 

[acmilan]

If it is a tangent then the line will only cut the parabola at one point. When you solve simultaneously you will therefore have equal roots. For equal roots, delta must = 0

 

[velox]

ah ok, i thought disriminant was only for roots with the x axis oops!

 

[acmilan]

Nope you can use it for any quadratic equation