A coplanar beam of light emerging from a point source have the equation λ x - y +2(1+λ)=0, ∀ λ ∈ R; the rays of the beam strike an elliptical surface and get reflected inside the ellipse. The reflected rays form another convergent beam having the equation μ x - y +2(1-μ)=0, ∀ μ ∈ R. Further it is found that the foot of the perpendicular from the point (2,2) upon any tangent to the ellipse lies on the circle x2+y2-4 y-5=0. The area of the largest triangle that an incident ray and corresponding reflected ray can enclose with major axis of the ellipse is equal to

Question

A coplanar beam of light emerging from a point source have the equation λ x - y +2(1+λ)=0, ∀ λ ∈ R; the rays of the beam strike an elliptical surface and get reflected inside the ellipse. The reflected rays form another convergent beam having the equation μ x - y +2(1-μ)=0, ∀ μ ∈ R. Further it is found that the foot of the perpendicular from the point (2,2) upon any tangent to the ellipse lies on the circle x2+y2-4 y-5=0. The area of the largest triangle that an incident ray and corresponding reflected ray can enclose with major axis of the ellipse is equal to (A) 4 √5 (B) √5 (C) 3 √5 (D) 2 √5

Tardigrade Admin · Accepted Answer

Area of Δ PF 1 F 2=(1/2) × base × height =(1/2) × 4 × height maximum are = maximum height = b ∵ b2=a2(1-e2)=9(1-(4/9)) ⇒ b=√5 ⇒ text Area =2 √5

$45$

$5$

$35$

$25$

Area of $Δ P F_{1} F_{2} = \frac{1}{2} \times$ base $\times$ height $= \frac{1}{2} \times 4 \times$ height maximum are $=$ maximum height $= b$
$∵ b^{2} = a^{2} (1 - e^{2}) = 9 (1 - \frac{4}{9})$
$\Rightarrow b = 5$
$\Rightarrow Area = 25$

45​

5​

35​

25​

Area of ΔPF1​F2​=21​× base × height =21​×4× height maximum are = maximum height =b ∵b2=a2(1−e2)=9(1−94​) ⇒b=5​ ⇒ Area =25​

$45$

$5$

$35$

$25$

Area of $Δ P F_{1} F_{2} = \frac{1}{2} \times$ base $\times$ height $= \frac{1}{2} \times 4 \times$ height maximum are $=$ maximum height $= b$
$∵ b^{2} = a^{2} (1 - e^{2}) = 9 (1 - \frac{4}{9})$
$\Rightarrow b = 5$
$\Rightarrow Area = 25$