Question

56. P is a point on the axis of the parabola y2 = 4ax;
Q and R are the extremities of its latus rectum, A
is its vertex. If PQR is an equilateral triangle lying
within the parabola and ‹AQP =(theta), then cos (theta) =
​

Answer 1

Given :   P is a point on the axis of the parabola y2 = 4ax;  Q and R are the extremities of its latus rectum, A is its vertex. If PQR is an equilateral triangle lying within the parabola

To find : Cos∠AQP

Solution:

PQR is an equilateral triangle

=> ∠RQP = 60°

in  Δ AQR

QR = 4a  

Let say AM  ⊥ QR

=> AM = a

QM = QR = 4a/2 = 2a

=>  Tan ∠AQR  =  a/2a

=> Tan ∠AQR = 1/2

or Cos∠AQR  = 2/√5   or Sin∠AQR = 1/√5

∠AQP  = ∠RQP  + ∠AQR

=> Cos∠AQP = Cos ( 60°  + ∠AQR)

=> Cos∠AQP = Cos60°Cos( ∠AQR)  - Sin60 Sin(∠AQR)

=>  Cos∠AQP =  (1/2)(2/√5)  -  (√3 /2)(1/√5 )

=> Cos∠AQP =  1/√5  - √3/2√5

=>  Cos∠AQP = (2 - √3)/2√5

=> Cos∠AQP = 0.06

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परवलय का समीकरण ज्ञात कीजिए जो दिए प्रतिबंध को ...

https://brainly.in/question/15777568