Q.175. Ellipse and normal

Question 175.
Let P be a point on the ellipse with parametric equations x = 5cost, y = 3sint for 0 < = t < 2π, and
let F and G be the points (-4,0) and (4,0) respectively. Prove that
(a) FP = 5 + 4cost,  
(b) FP + PG = 10.
Let the normal at P make angles "θ" and "Φ" with FP and GP respectively. Prove that
(c) tanθ = 4/3 sint
(d) θ = Φ.

Answer 175.
(a)
FP^2 (using distance formula)
= (5cost +4)^2 + (3sint)^2
= 25cos^2 t + 40cost + 16 + 9 sin^2 t
= 25cos^2 t + 40cost + 16 + 9(1 - cos^2 t)
= 16cos^2 t + 40cost + 25
= (5 + 4cost)^2
=> FP = 5 + 4cost.

(b)
PG^2
= (5cost - 4)^2 + (3sint)^2
= 25cos^2 t - 40cost + 16 + 9(1 - cos^2 t)
= 16cos^2 t - 40cost + 25
= (5 - 4cost)^2
=> PG = 5 - 4cost
=> FP + PG = (5 + 4cost) + (5 - 4cost) = 10.

(c)
x = 5cost and y = 3sint
=> dx/dt = -5sint and dy/dt = 3cost
=> slope of tangent at p = dy/dx
= (dy/dt)/(dx/dt)
= - (3/5) cot t
=> slope of normal at P = (5/3) tant = m1
Slope of FP
= 3sint/(5cost + 4) = m2
Angle between normal and FP is given by
tanθ
= (m1 - m2)/(1 + m1m2)
= [(5/3)tant - 3sint/(5cost + 4)] / [1 + 5tant sint / (5cost + 4)]
= (1/3)sint*[5/cost - 9/(5cost + 4)] / [1 + 5sin^2t / (5cos^2 t + 4cost)]
= (1/3)sint * (25cost + 20 - 9cost) / (5cos^2 t + 4cost + 5sin^2 t)
= (4/3) sint * (4cost + 5) / (5 + 4cost)
= (4/3) sint.

(d)
Slope of GP
= 3sint/(5cost - 4) = m3
Angle between normal and GP is given by
tanΦ
= l (m1 - m3)/(1 + m1m3) l
= l [(5/3)tant - 3sint/(5cost - 4)] / [1 + 5tant sint / (5cost - 4)] l
= (1/3)sint*l [5/cost - 9/(5cost - 4)] / [1 + 5sin^2t / (5cos^2 t - 4cost)] l
= (1/3)sint * l (25cost - 20 - 9cost) / (5cos^2 t - 4cost + 5sin^2 t) l
= (4/3) sint * l (4cost - 5) / (5 - 4cost) l
= (4/3) sint.
=> tanθ = tanΦ
=> θ = Φ.

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