Sum hard questions (1 Viewer)

[GaganDeep]

Here are some questions i need help wif. thx

 

[bboyelement]

for F draw up your diagram according to the description.

vector PR = z₁ - z₃
vector PQ = z₁ - z₂

using pythagoras.

PR² + PQ² = a²
z₁² - 2z₁z₃ + z₃² + z₁² - 2z₁z₂ + z₂² = a²

but we know that vector PQ = vector PR x i (as your are turning 90 degrees)

therefore PR² + (i PR)²
= PR² - PR² = 0
therefore a = 0

and from the equation above you should get the answer by moving it to the other side and adding z1

 

[GaganDeep]

ok, thanks, what bout hte rest

 

[gamecw]

For the ellipse question

let P be (2cos@, rt3sin@)

equation of normal at P: y=-2/rt3.tan@.x - [(4-3)/rt3]sin@

let C be the pt of intersection of normal n X axis

Solve for C get: x= [(4-3)/2]cos@

now use that to prove PS/PS'=CS/CS' (shud have done that in previous part?)

Now finally:
let ^PCS=q (an angle)
^PCS'=180-q

use sine rule sin(p)/PS=sin(^CPS)/CS and sin(180-p)/PS'=sin(^CPS')/CS'

since sin(p)=sin(180-p) and PS/PS'=CS/CS', derive ^CPS=^CPS'