help with a past trial question (1 Viewer)

[iheartgeorge]

can anybody explain how to do this question please. it's from a past paper that i dont have the answers for

(a) (i) Prove that the line with equation y = px + (1 - 2p^2) is a tangent to the parabola x^2 = 8 (y-1) for all values of p

(ii) Find the angle between the tangents drawn to x^2=8(y-1) from the point (0,-7)

 

[Horseypie]

a) i) from x^2 = 8(y-1) make y the subject then solve both equations simultaneously...you should get one value for x which means its a tangent

ii) find the equations of the tangents using normal calculus and then use the angle between two lines formula

 

[Forbidden.]

a) i) from x^2 = 8(y-1) make y the subject then solve both equations simultaneously...you should get one value for x which means its a tangent

ii) find the equations of the tangents using normal calculus and then use the angle between two lines formula

If you mean tan θ = |m₁-m₂/1+m₁m₂| for ii) I wouldn't think of using it in 2U

 

[luthrax]

dont you have to use the discriminate

 

[luthrax]

once u solve simultaneously u get x^2 -8px + 16p^2
the disriminate=( 64p^2 -64p^2)0.5

therefore for all values of p discrimate is equal to zero, therefore a tangent at all values of p

please help me with my question http://community.boredofstudies.org/12/mathematics/155617/2-unit-geometry.html

lol

 

[luthrax]

for part ii, as the point is vertically below the vertex of the parabola, the angle that forms between the tangents is twice the angle of the gradient of one of the tangents, draw a rough sketch u'll get wat im saying, since tan x=m, 2x gives u the angle between the two tangents.
you find the tangents by subbing (0, -7) into y=px+ (1 +2p^2)

 

[Horseypie]

If you mean tan θ = |m₁-m₂/1+m₁m₂| for ii) I wouldn't think of using it in 2U

Oh is that 3 unit?...sorry, i didnt think about that.