x 2 4 + y 2 3 = 1 \dfrac{x^2}{4} + \dfrac{y^2}{3} = 1 4 x 2 + 3 y 2 = 1
vertical line x = h meets ellipse at P , Q \text{vertical line } x = h \text{ meets ellipse at } P, Q vertical line x = h meets ellipse at P , Q
tangents at P , Q meet at R \text{tangents at } P, Q \text{ meet at } R tangents at P , Q meet at R
Δ(h) = area of triangle PQR
Δ 1 = max 1 / 2 ≤ h ≤ 1 Δ ( h ) , Δ 2 = min 1 / 2 ≤ h ≤ 1 Δ ( h ) \Delta_1 = \max_{1/2 \le h \le 1}\Delta(h),\quad \Delta_2 = \min_{1/2 \le h \le 1}\Delta(h) Δ 1 = max 1/2 ≤ h ≤ 1 Δ ( h ) , Δ 2 = min 1/2 ≤ h ≤ 1 Δ ( h )
find 8 5 Δ 1 − 8 Δ 2 \text{find } \dfrac{8}{\sqrt{5}}\Delta_1 - 8\Delta_2 find 5 8 Δ 1 − 8 Δ 2
1. coordinates of P , Q 1.\ \text{coordinates of } P, Q 1. coordinates of P , Q
2. tangents → R 2.\ \text{tangents} \to R 2. tangents → R
3. Δ ( h ) → max/min on [ 1 2 , 1 ] 3.\ \Delta(h) \to \text{max/min on } \left[\tfrac{1}{2}, 1\right] 3. Δ ( h ) → max/min on [ 2 1 , 1 ]
x = h ⇒ h 2 4 + y 2 3 = 1 x = h \Rightarrow \dfrac{h^2}{4} + \dfrac{y^2}{3} = 1 x = h ⇒ 4 h 2 + 3 y 2 = 1
y 2 = 3 ( 4 − h 2 ) 4 ⇒ y = ± 3 2 4 − h 2 y^2 = \dfrac{3(4-h^2)}{4} \Rightarrow y = \pm\dfrac{\sqrt{3}}{2}\sqrt{4-h^2} y 2 = 4 3 ( 4 − h 2 ) ⇒ y = ± 2 3 4 − h 2
P ( h , 3 2 4 − h 2 ) , Q ( h , − 3 2 4 − h 2 ) P\!\left(h, \dfrac{\sqrt{3}}{2}\sqrt{4-h^2}\right),\ Q\!\left(h, -\dfrac{\sqrt{3}}{2}\sqrt{4-h^2}\right) P ( h , 2 3 4 − h 2 ) , Q ( h , − 2 3 4 − h 2 )
tangent at ( x 1 , y 1 ) : x x 1 4 + y y 1 3 = 1 \text{tangent at } (x_1,y_1):\ \dfrac{x x_1}{4} + \dfrac{y y_1}{3} = 1 tangent at ( x 1 , y 1 ) : 4 x x 1 + 3 y y 1 = 1
at P : h x 4 + y 0 y 3 = 1 \text{at } P:\ \dfrac{hx}{4} + \dfrac{y_0 y}{3} = 1 at P : 4 h x + 3 y 0 y = 1
at Q : h x 4 − y 0 y 3 = 1 \text{at } Q:\ \dfrac{hx}{4} - \dfrac{y_0 y}{3} = 1 at Q : 4 h x − 3 y 0 y = 1
add : 2 h x 4 = 2 ⇒ x = 4 h \text{add}:\ \dfrac{2hx}{4} = 2 \Rightarrow x = \dfrac{4}{h} add : 4 2 h x = 2 ⇒ x = h 4
subtract : 2 y 0 y 3 = 0 ⇒ y = 0 \text{subtract}:\ \dfrac{2y_0 y}{3} = 0 \Rightarrow y = 0 subtract : 3 2 y 0 y = 0 ⇒ y = 0
R ( 4 h , 0 ) R\!\left(\dfrac{4}{h},\ 0\right) R ( h 4 , 0 )
P Q = 2 y 0 = 3 4 − h 2 PQ = 2y_0 = \sqrt{3}\,\sqrt{4-h^2} P Q = 2 y 0 = 3 4 − h 2
height from R to x = h : 4 h − h = 4 − h 2 h \text{height from } R \text{ to } x = h:\ \dfrac{4}{h} - h = \dfrac{4-h^2}{h} height from R to x = h : h 4 − h = h 4 − h 2
Δ ( h ) = 1 2 ( P Q ) ( height ) = 1 2 ⋅ 3 4 − h 2 ⋅ 4 − h 2 h \Delta(h) = \tfrac{1}{2}(PQ)(\text{height}) = \tfrac{1}{2}\cdot\sqrt{3}\sqrt{4-h^2}\cdot\dfrac{4-h^2}{h} Δ ( h ) = 2 1 ( P Q ) ( height ) = 2 1 ⋅ 3 4 − h 2 ⋅ h 4 − h 2
= 3 2 ⋅ ( 4 − h 2 ) 3 / 2 h = \dfrac{\sqrt{3}}{2}\cdot\dfrac{(4-h^2)^{3/2}}{h} = 2 3 ⋅ h ( 4 − h 2 ) 3/2
f ( h ) = ( 4 − h 2 ) 3 / 2 h f(h) = \dfrac{(4-h^2)^{3/2}}{h} f ( h ) = h ( 4 − h 2 ) 3/2
f ′ ( h ) = − 2 4 − h 2 ( h 2 + 2 ) h 2 < 0 f'(h) = \dfrac{-2\sqrt{4-h^2}\,(h^2+2)}{h^2} < 0 f ′ ( h ) = h 2 − 2 4 − h 2 ( h 2 + 2 ) < 0
⇒ Δ decreasing on [ 1 2 , 1 ] \Rightarrow \Delta \text{ decreasing on } \left[\tfrac{1}{2}, 1\right] ⇒ Δ decreasing on [ 2 1 , 1 ]
Δ 1 = Δ ( 1 2 ) = 3 2 ⋅ ( 15 4 ) 3 / 2 1 2 \Delta_1 = \Delta\!\left(\tfrac{1}{2}\right) = \dfrac{\sqrt{3}}{2}\cdot\dfrac{\left(\tfrac{15}{4}\right)^{3/2}}{\tfrac{1}{2}} Δ 1 = Δ ( 2 1 ) = 2 3 ⋅ 2 1 ( 4 15 ) 3/2
= 3 2 ⋅ 2 ⋅ 15 15 8 = 45 5 8 = \dfrac{\sqrt{3}}{2}\cdot 2\cdot\dfrac{15\sqrt{15}}{8} = \dfrac{45\sqrt{5}}{8} = 2 3 ⋅ 2 ⋅ 8 15 15 = 8 45 5
Δ 2 = Δ ( 1 ) = 3 2 ⋅ 3 3 / 2 = 9 2 \Delta_2 = \Delta(1) = \dfrac{\sqrt{3}}{2}\cdot 3^{3/2} = \dfrac{9}{2} Δ 2 = Δ ( 1 ) = 2 3 ⋅ 3 3/2 = 2 9
8 5 Δ 1 − 8 Δ 2 = 8 5 ⋅ 45 5 8 − 8 ⋅ 9 2 \dfrac{8}{\sqrt{5}}\Delta_1 - 8\Delta_2 = \dfrac{8}{\sqrt{5}}\cdot\dfrac{45\sqrt{5}}{8} - 8\cdot\dfrac{9}{2} 5 8 Δ 1 − 8 Δ 2 = 5 8 ⋅ 8 45 5 − 8 ⋅ 2 9
= 45 − 36 = 9 = 45 - 36 = 9 = 45 − 36 = 9
8 5 Δ 1 − 8 Δ 2 = 9 \dfrac{8}{\sqrt{5}}\Delta_1 - 8\Delta_2 = 9 5 8 Δ 1 − 8 Δ 2 = 9
9 \boxed{9} 9
