Question-89849

Question Number 89849 by ajfour last updated on 19/Apr/20

Commented by ajfour last updated on 19/Apr/20

Find maximum area of △PQR in terms of ellipse parameters a,b. or find side of △PQR when it is equilateral.

F i n d m a x i m u m a r e a o f △ P Q R

i n t e r m s o f e l l i p s e p a r a m e t e r s a, b .

o r f i n d s i d e o f △ P Q R w h e n

i t i s e q u i l a t e r a l .

Answered by mr W last updated on 19/Apr/20

PART I r=b P ′(r cos θ, r sin θ) Q ′(r sin θ, r cos θ) A=(r^2 /2)tan ((π/4)+(θ/2))(cos θ−sin θ)^2 A=(r^2 /2)×((1+tan (θ/2))/(1−tan (θ/2)))(((1−tan^2 (θ/2)−2 tan (θ/2))/(1+tan^2 (θ/2))))^2 let t=tan (θ/2) ⇒((2A)/r^2 )=((1+t)/(1−t))(((1−t^2 −2t)/(1+t^2 )))^2 (d/dt)(((2A)/r^2 ))=0 ..... ⇒t=−0.1526 ⇒θ=−17.3528° A_(max) =((1.1538r^2 )/2)=0.5769r^2 =0.5769ab

P A R T I

r = b

P^{'} (r \cos θ, r \sin θ)

Q^{'} (r \sin θ, r \cos θ)

A = \frac{r^{2}}{2} \tan (\frac{π}{4} + \frac{θ}{2}) {(\cos θ - \sin θ)}^{2}

A = \frac{r^{2}}{2} \times \frac{1 + \tan \frac{θ}{2}}{1 - \tan \frac{θ}{2}} {(\frac{1 - \tan^{2} \frac{θ}{2} - 2 \tan \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}})}^{2}

l e t t = \tan \frac{θ}{2}

\Rightarrow \frac{2 A}{r^{2}} = \frac{1 + t}{1 - t} {(\frac{1 - t^{2} - 2 t}{1 + t^{2}})}^{2}

\frac{d}{d t} (\frac{2 A}{r^{2}}) = 0

\dots . .

\Rightarrow t = - 0.1526

\Rightarrow θ = - 17.3528 °

A_{m a x} = \frac{1.1538 r^{2}}{2} = 0.5769 r^{2} = 0.5769 a b

Commented by mr W last updated on 19/Apr/20

Commented by ajfour last updated on 20/Apr/20

sir why x_(P′) =y_(Q′) & y_(P′) =x_(Q′) and where does ′a′ go?

s i r w h y x_{P^{'}} = y_{Q^{'}} & y_{P^{'}} = x_{Q^{'}}

a n d w h e r e d o e s^{'} a^{'} g o ?

Commented by mr W last updated on 20/Apr/20

in case of circle we use the symmetry for the max. area. in case of ellipse we only need to stretch the circle in x−direction in ratio (a/r), i.e. (a/b).

i n c a s e o f c i r c l e w e u s e t h e s y m m e t r y

f o r t h e m a x . a r e a .

i n c a s e o f e l l i p s e w e o n l y n e e d t o

s t r e t c h t h e c i r c l e i n x - d i r e c t i o n i n

r a t i o \frac{a}{r}, i . e . \frac{a}{b} .

Answered by mr W last updated on 19/Apr/20

Commented by mr W last updated on 19/Apr/20

PART II the point R must lie on the green circle with center at C. let α=tan^(−1) (b/a), μ=(b/a) AC=((AB)/(2 cos 30°))=(√((a^2 +b^2 )/3)) x_C =−a+AC cos (30°−α) x_C =−a+(√((a^2 +b^2 )/3)) (((√3)/2)×(a/( (√(a^2 +b^2 ))))+(1/2)×(b/( (√(a^2 +b^2 ))))) ⇒x_C =−(1/2) (a−(b/( (√3)))) y_C =AC sin (30°−α)=(√((a^2 +b^2 )/3))((1/2)×(a/( (√(a^2 +b^2 ))))−((√3)/2)×(b/( (√(a^2 +b^2 ))))) ⇒y_C =(1/2)((a/( (√3)))−b) eqn. of green circle: [x+(1/2)(a−(b/( (√3))))]^2 +[y−(1/2)((a/( (√3)))−b)]^2 =((a^2 +b^2 )/3) or x_R =−(1/2)(a−(b/( (√3))))+(√((a^2 +b^2 )/3)) cos θ y_R =(1/2)((a/( (√3)))−b)+(√((a^2 +b^2 )/3)) sin θ ⇒(x_R /a)=−(1/2)(1−(μ/( (√3))))+(√((1+μ^2 )/3)) cos θ ⇒(y_R /a)=(1/2)((1/( (√3)))−μ)+(√((1+μ^2 )/3)) sin θ eqn. of AR: y=((a−(√3)b+2(√(a^2 +b^2 )) sin θ)/( (√3)a+b+2(√(a^2 +b^2 )) cos θ))(x+a) y=((1−(√3)μ+2(√(1+μ^2 )) sin θ)/( (√3)+μ+2(√(1+μ^2 )) cos θ))(x+a) intersection point P: y_P =((1−(√3)μ+2(√(1+μ^2 )) sin θ)/( (√3)+μ+2(√(1+μ^2 )) cos θ))(x_P +a) ((x_P /a))^2 +((y_P /b))^2 =1 (((1−(√3)μ+2(√(1+μ^2 )) sin θ)/( (√3)+μ+2(√(1+μ^2 )) cos θ)))^2 (1+(x_P /a))^2 =μ^2 [1−((x_P /a))^2 ] (((1−(√3)μ+2(√(1+μ^2 )) sin θ)/( (√3)+μ+2(√(1+μ^2 )) cos θ)))^2 (1+(x_P /a))=μ^2 (1−(x_P /a)) ⇒(x_P /a)=((μ^2 −(((1−(√3)μ+2(√(1+μ^2 )) sin θ)/( (√3)+μ+2(√(1+μ^2 )) cos θ)))^2 )/(μ^2 +(((1−(√3)μ+2(√(1+μ^2 )) sin θ)/( (√3)+μ+2(√(1+μ^2 )) cos θ)))^2 )) ⇒(y_P /a)=(((1−(√3)μ+2(√(1+μ^2 )) sin θ)/( (√3)+μ+2(√(1+μ^2 )) cos θ)))(1+(x_P /a)) ((PR^2 )/a^2 )=((x_P /a)−(x_R /a))^2 +((y_P /a)−(y_R /a))^2 eqn. of BR: ((y+b)/(y_R +b))=(x/x_R ) y=−b+((a+(√3)b+2(√(a^2 +b^2 )) sin θ)/(−(√3)a+b+2(√(a^2 +b^2 )) cos θ))x y=−b+((1+(√3)μ+2(√(1+μ^2 )) sin θ)/(−(√3)+μ+2(√(1+μ^2 )) cos θ))x intersection point Q: y_Q =−b+((1+(√3)μ+2(√(1+μ^2 )) sin θ)/(−(√3)+μ+2(√(1+μ^2 )) cos θ))x_Q x_Q =((−(√3)+μ+2(√(1+μ^2 )) cos θ)/(1+(√3)μ+2(√(1+μ^2 )) sin θ))(y_Q +b) ((x_Q /a))^2 +((y_Q /b))^2 =1 μ^2 (((−(√3)+μ+2(√(1+μ^2 )) cos θ)/(1+(√3)μ+2(√(1+μ^2 )) sin θ)))^2 ((y_Q /a)+μ)^2 +((y_Q /a))^2 =μ^2 μ^2 (((−(√3)+μ+2(√(1+μ^2 )) cos θ)/(1+(√3)μ+2(√(1+μ^2 )) sin θ)))^2 ((y_Q /a)+μ)=μ−(y_Q /a) ⇒(y_Q /a)=((μ[1−μ^2 (((−(√3)+μ+2(√(1+μ^2 )) cos θ)/(1+(√3)μ+2(√(1+μ^2 )) sin θ)))^2 ])/(1+μ^2 (((−(√3)+μ+2(√(1+μ^2 )) cos θ)/(1+(√3)μ+2(√(1+μ^2 )) sin θ)))^2 )) ⇒(x_Q /a)=(((−(√3)+μ+2(√(1+μ^2 )) cos θ)/(1+(√3)μ+2(√(1+μ^2 )) sin θ)))((y_Q /a)+μ) ((QR^2 )/a^2 )=((x_Q /a)−(x_R /a))^2 +((y_Q /a)−(y_R /a))^2 such that ΔPQR is equilateral, PR=QR ((PR^2 )/a^2 )=((QR^2 )/a^2 )

P A R T I I

t h e p o i n t R m u s t l i e o n t h e g r e e n c i r c l e

w i t h c e n t e r a t C .

l e t α = \tan^{- 1} \frac{b}{a}, μ = \frac{b}{a}

A C = \frac{A B}{2 \cos 30 °} = \sqrt{\frac{a^{2} + b^{2}}{3}}

x_{C} = - a + A C \cos (30 ° - α)

x_{C} = - a + \sqrt{\frac{a^{2} + b^{2}}{3}} (\frac{\sqrt{3}}{2} \times \frac{a}{\sqrt{a^{2} + b^{2}}} + \frac{1}{2} \times \frac{b}{\sqrt{a^{2} + b^{2}}})

\Rightarrow x_{C} = - \frac{1}{2} (a - \frac{b}{\sqrt{3}})

y_{C} = A C \sin (30 ° - α) = \sqrt{\frac{a^{2} + b^{2}}{3}} (\frac{1}{2} \times \frac{a}{\sqrt{a^{2} + b^{2}}} - \frac{\sqrt{3}}{2} \times \frac{b}{\sqrt{a^{2} + b^{2}}})

\Rightarrow y_{C} = \frac{1}{2} (\frac{a}{\sqrt{3}} - b)

e q n . o f g r e e n c i r c l e :

{[x + \frac{1}{2} (a - \frac{b}{\sqrt{3}})]}^{2} + {[y - \frac{1}{2} (\frac{a}{\sqrt{3}} - b)]}^{2} = \frac{a^{2} + b^{2}}{3}

o r

x_{R} = - \frac{1}{2} (a - \frac{b}{\sqrt{3}}) + \sqrt{\frac{a^{2} + b^{2}}{3}} \cos θ

y_{R} = \frac{1}{2} (\frac{a}{\sqrt{3}} - b) + \sqrt{\frac{a^{2} + b^{2}}{3}} \sin θ

\Rightarrow \frac{x_{R}}{a} = - \frac{1}{2} (1 - \frac{μ}{\sqrt{3}}) + \sqrt{\frac{1 + μ^{2}}{3}} \cos θ

\Rightarrow \frac{y_{R}}{a} = \frac{1}{2} (\frac{1}{\sqrt{3}} - μ) + \sqrt{\frac{1 + μ^{2}}{3}} \sin θ

e q n . o f A R :

y = \frac{a - \sqrt{3} b + 2 \sqrt{a^{2} + b^{2}} \sin θ}{\sqrt{3} a + b + 2 \sqrt{a^{2} + b^{2}} \cos θ} (x + a)

y = \frac{1 - \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ}{\sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ} (x + a)

i n t e r s e c t i o n p o i n t P :

y_{P} = \frac{1 - \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ}{\sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ} (x_{P} + a)

{(\frac{x_{P}}{a})}^{2} + {(\frac{y_{P}}{b})}^{2} = 1

{(\frac{1 - \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ}{\sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ})}^{2} {(1 + \frac{x_{P}}{a})}^{2} = μ^{2} [1 - {(\frac{x_{P}}{a})}^{2}]

{(\frac{1 - \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ}{\sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ})}^{2} (1 + \frac{x_{P}}{a}) = μ^{2} (1 - \frac{x_{P}}{a})

\Rightarrow \frac{x_{P}}{a} = \frac{μ^{2} - {(\frac{1 - \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ}{\sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ})}^{2}}{μ^{2} + {(\frac{1 - \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ}{\sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ})}^{2}}

\Rightarrow \frac{y_{P}}{a} = (\frac{1 - \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ}{\sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ}) (1 + \frac{x_{P}}{a})

\frac{{P R}^{2}}{a^{2}} = {(\frac{x_{P}}{a} - \frac{x_{R}}{a})}^{2} + {(\frac{y_{P}}{a} - \frac{y_{R}}{a})}^{2}

e q n . o f B R :

\frac{y + b}{y_{R} + b} = \frac{x}{x_{R}}

y = - b + \frac{a + \sqrt{3} b + 2 \sqrt{a^{2} + b^{2}} \sin θ}{- \sqrt{3} a + b + 2 \sqrt{a^{2} + b^{2}} \cos θ} x

y = - b + \frac{1 + \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ}{- \sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ} x

i n t e r s e c t i o n p o i n t Q :

y_{Q} = - b + \frac{1 + \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ}{- \sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ} x_{Q}

x_{Q} = \frac{- \sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ}{1 + \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ} (y_{Q} + b)

{(\frac{x_{Q}}{a})}^{2} + {(\frac{y_{Q}}{b})}^{2} = 1

μ^{2} {(\frac{- \sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ}{1 + \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ})}^{2} {(\frac{y_{Q}}{a} + μ)}^{2} + {(\frac{y_{Q}}{a})}^{2} = μ^{2}

μ^{2} {(\frac{- \sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ}{1 + \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ})}^{2} (\frac{y_{Q}}{a} + μ) = μ - \frac{y_{Q}}{a}

\Rightarrow \frac{y_{Q}}{a} = \frac{μ [1 - μ^{2} {(\frac{- \sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ}{1 + \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ})}^{2}]}{1 + μ^{2} {(\frac{- \sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ}{1 + \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ})}^{2}}

\Rightarrow \frac{x_{Q}}{a} = (\frac{- \sqrt{3} + μ + 2 \sqrt{1 + μ^{2}} \cos θ}{1 + \sqrt{3} μ + 2 \sqrt{1 + μ^{2}} \sin θ}) (\frac{y_{Q}}{a} + μ)

\frac{{Q R}^{2}}{a^{2}} = {(\frac{x_{Q}}{a} - \frac{x_{R}}{a})}^{2} + {(\frac{y_{Q}}{a} - \frac{y_{R}}{a})}^{2}

s u c h t h a t Δ P Q R i s e q u i l a t e r a l,

P R = Q R

\frac{{P R}^{2}}{a^{2}} = \frac{{Q R}^{2}}{a^{2}}

Commented by mr W last updated on 19/Apr/20

((x_P /a)−(x_R /a))^2 +((y_P /a)−(y_R /a))^2 =((x_Q /a)−(x_R /a))^2 +((y_Q /a)−(y_R /a))^2 from this eqn. we get θ for the position of point R.

{(\frac{x_{P}}{a} - \frac{x_{R}}{a})}^{2} + {(\frac{y_{P}}{a} - \frac{y_{R}}{a})}^{2} = {(\frac{x_{Q}}{a} - \frac{x_{R}}{a})}^{2} + {(\frac{y_{Q}}{a} - \frac{y_{R}}{a})}^{2}

f r o m t h i s e q n . w e g e t θ f o r t h e

p o s i t i o n o f p o i n t R .

Commented by mr W last updated on 19/Apr/20

Commented by mr W last updated on 20/Apr/20

Commented by mr W last updated on 20/Apr/20

Commented by mr W last updated on 20/Apr/20

Commented by mr W last updated on 21/Apr/20

Answered by ajfour last updated on 20/Apr/20

Commented by ajfour last updated on 20/Apr/20

let R(h,k) Eq. AP: y=((k(x+a))/(h+a)) b^2 x^2 +((k^2 (x+a)^2 a^2 )/((h+a)^2 ))=a^2 b^2 (h+a)^2 b^2 x^2 +k^2 a^2 (x+a)^2 =a^2 b^2 (h+a)^2 roots are x_P , −a ⇒ −ax_P = ((a^2 [a^2 k^2 −b^2 (h+a)^2 ])/(b^2 (h+a)^2 +k^2 a^2 )) x_P =((a[b^2 (h+a)^2 −a^2 k^2 ])/(b^2 (h+a)^2 +k^2 a^2 )) ...(I) Eq. of BQ: y=(((k+b)/h))x−b b^2 h^2 x^2 +a^2 [(k+b)x−bh]^2 =a^2 b^2 h^2 roots are x=0, x_Q x_Q = ((2a^2 bh(k+b))/(b^2 h^2 +a^2 (k+b)^2 )) ...(II) ((((k+b)/h)−(k/(h+a)))/(1+(((k+b)/h))((k/(h+a)))))=(√3) ....(III) slope of PQ: ((y_Q −y_P )/(x_P −x_Q ))=(((√3)−(k/(h+a)))/(1+(√3)((k/(h+a))))) ...(IV) let (k/(h+a))=p , ((k+b)/h)=q x_P =((a[b^2 (h+a)^2 −a^2 k^2 ])/(b^2 (h+a)^2 +k^2 a^2 )) = ((a(b^2 −a^2 p^2 ))/(b^2 +a^2 p^2 )) x_Q =((2a^2 bh(k+b))/(b^2 h^2 +a^2 (k+b)^2 )) = ((2a^2 bq)/(b^2 +a^2 q^2 )) y_P =p(x_P +a)=((p(2ab^2 ))/(b^2 +a^2 p^2 )) y_Q =qx_Q −b = ((2a^2 bq^2 )/(b^2 +a^2 q^2 ))−b =((b(a^2 q^2 −b^2 ))/(b^2 +a^2 q^2 )) ((((b(a^2 q^2 −b^2 ))/(b^2 +a^2 q^2 ))−((p(2ab^2 ))/(b^2 +a^2 p^2 )))/(((a(b^2 −a^2 p^2 ))/(b^2 +a^2 p^2 ))−((2a^2 bq)/(b^2 +a^2 q^2 ))))=(((√3)−p)/(1+p(√3))) ......(i) And ((q−p)/(1+pq))=(√3) .......(ii) ((b(a^2 q^2 −b^2 )(b^2 +a^2 p^2 )−2ab^2 p(b^2 +a^2 q^2 ))/(a(b^2 −a^2 p^2 )(b^2 +a^2 q^2 )+2a^2 bq(b^2 +a^2 p^2 ))) = (((√3)−p)/(1+p(√3))) ....

l e t R (h, k)

E q . A P : y = \frac{k (x + a)}{h + a}

b^{2} x^{2} + \frac{k^{2} {(x + a)}^{2} a^{2}}{{(h + a)}^{2}} = a^{2} b^{2}

{(h + a)}^{2} b^{2} x^{2} + k^{2} a^{2} {(x + a)}^{2} = a^{2} b^{2} {(h + a)}^{2}

r o o t s a r e x_{P}, - a

\Rightarrow - {a x}_{P} = \frac{a^{2} [a^{2} k^{2} - b^{2} {(h + a)}^{2}]}{b^{2} {(h + a)}^{2} + k^{2} a^{2}}

x_{P} = \frac{a [b^{2} {(h + a)}^{2} - a^{2} k^{2}]}{b^{2} {(h + a)}^{2} + k^{2} a^{2}} \dots (I)

E q . o f B Q :

y = (\frac{k + b}{h}) x - b

b^{2} h^{2} x^{2} + a^{2} {[(k + b) x - b h]}^{2} = a^{2} b^{2} h^{2}

r o o t s a r e x = 0, x_{Q}

x_{Q} = \frac{2 a^{2} b h (k + b)}{b^{2} h^{2} + a^{2} {(k + b)}^{2}} \dots (I I)

\frac{\frac{k + b}{h} - \frac{k}{h + a}}{1 + (\frac{k + b}{h}) (\frac{k}{h + a})} = \sqrt{3} \dots . (I I I)

s l o p e o f P Q :

\frac{y_{Q} - y_{P}}{x_{P} - x_{Q}} = \frac{\sqrt{3} - \frac{k}{h + a}}{1 + \sqrt{3} (\frac{k}{h + a})} \dots (I V)

l e t \frac{k}{h + a} = p, \frac{k + b}{h} = q

x_{P} = \frac{a [b^{2} {(h + a)}^{2} - a^{2} k^{2}]}{b^{2} {(h + a)}^{2} + k^{2} a^{2}}

= \frac{a (b^{2} - a^{2} p^{2})}{b^{2} + a^{2} p^{2}}

x_{Q} = \frac{2 a^{2} b h (k + b)}{b^{2} h^{2} + a^{2} {(k + b)}^{2}}

= \frac{2 a^{2} b q}{b^{2} + a^{2} q^{2}}

y_{P} = p (x_{P} + a) = \frac{p (2 {a b}^{2})}{b^{2} + a^{2} p^{2}}

y_{Q} = {q x}_{Q} - b = \frac{2 a^{2} {b q}^{2}}{b^{2} + a^{2} q^{2}} - b

= \frac{b (a^{2} q^{2} - b^{2})}{b^{2} + a^{2} q^{2}}

\frac{\frac{b (a^{2} q^{2} - b^{2})}{b^{2} + a^{2} q^{2}} - \frac{p (2 {a b}^{2})}{b^{2} + a^{2} p^{2}}}{\frac{a (b^{2} - a^{2} p^{2})}{b^{2} + a^{2} p^{2}} - \frac{2 a^{2} b q}{b^{2} + a^{2} q^{2}}} = \frac{\sqrt{3} - p}{1 + p \sqrt{3}}

\dots \dots (i)

A n d \frac{q - p}{1 + p q} = \sqrt{3} \dots \dots . (i i)

\frac{b (a^{2} q^{2} - b^{2}) (b^{2} + a^{2} p^{2}) - 2 {a b}^{2} p (b^{2} + a^{2} q^{2})}{a (b^{2} - a^{2} p^{2}) (b^{2} + a^{2} q^{2}) + 2 a^{2} b q (b^{2} + a^{2} p^{2})}

= \frac{\sqrt{3} - p}{1 + p \sqrt{3}}

\dots .

Commented by mr W last updated on 20/Apr/20

you can get also a final eqn. with one unknown in this way.

y o u c a n g e t a l s o a f i n a l e q n . w i t h o n e

u n k n o w n i n t h i s w a y .

Commented by ajfour last updated on 21/Apr/20

I have a better way too, Sir, but lets leave it here; you solved it wonderfully, both parts, Thanks a lot!

I h a v e a b e t t e r w a y t o o, S i r,

b u t l e t s l e a v e i t h e r e; y o u

s o l v e d i t w o n d e r f u l l y, b o t h

p a r t s, T h a n k s a l o t!

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