← 2019 Paper 1
UPSC 2019 Maths Optional Paper 1 Q4b — Step-by-Step Solution 15 marks · Section A
Ellipsoid · Analytic Geometry · asked 3× in 13 yrs · Read the full method →
Question
Find the length of the normal chord through a point P P P x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1 a 2 x 2 + b 2 y 2 + c 2 z 2 = 1 4 P G 3 4PG_3 4 P G 3 G 3 G_3 G 3 P P P x y xy x y P P P x 2 a 6 ( 2 c 2 − a 2 ) + y 2 b 6 ( 2 c 2 − b 2 ) + z 2 c 4 = 0. \dfrac{x^2}{a^6}(2c^2-a^2)+\dfrac{y^2}{b^6}(2c^2-b^2)+\dfrac{z^2}{c^4}=0. a 6 x 2 ( 2 c 2 − a 2 ) + b 6 y 2 ( 2 c 2 − b 2 ) + c 4 z 2 = 0.
Technique
Parametrize the normal by arclength-multiple k k k k 2 k_2 k 2 z = 0 z=0 z = 0 k G 3 = − c 2 k_{G_3}=-c^2 k G 3 = − c 2 S \sqrt S S P P ′ = 4 P G 3 PP'=4PG_3 P P ′ = 4 P G 3 S = 2 c 2 S ′ S=2c^2S' S = 2 c 2 S ′
Solution
Let P = ( x 1 , y 1 , z 1 ) P=(x_1,y_1,z_1) P = ( x 1 , y 1 , z 1 )
n = ( x 1 a 2 , y 1 b 2 , z 1 c 2 ) . \mathbf n=\left(\frac{x_1}{a^2},\frac{y_1}{b^2},\frac{z_1}{c^2}\right). n = ( a 2 x 1 , b 2 y 1 , c 2 z 1 ) . Write S = x 1 2 a 4 + y 1 2 b 4 + z 1 2 c 4 = ∣ n ∣ 2 S=\dfrac{x_1^2}{a^4}+\dfrac{y_1^2}{b^4}+\dfrac{z_1^2}{c^4}=|\mathbf n|^2 S = a 4 x 1 2 + b 4 y 1 2 + c 4 z 1 2 = ∣ n ∣ 2 S ′ = x 1 2 a 6 + y 1 2 b 6 + z 1 2 c 6 S'=\dfrac{x_1^2}{a^6}+\dfrac{y_1^2}{b^6}+\dfrac{z_1^2}{c^6} S ′ = a 6 x 1 2 + b 6 y 1 2 + c 6 z 1 2
Step 1 — Parametrize the normal chord
Points on the normal: ( x 1 + x 1 a 2 k , y 1 + y 1 b 2 k , z 1 + z 1 c 2 k ) \big(x_1+\tfrac{x_1}{a^2}k,\ y_1+\tfrac{y_1}{b^2}k,\ z_1+\tfrac{z_1}{c^2}k\big) ( x 1 + a 2 x 1 k , y 1 + b 2 y 1 k , z 1 + c 2 z 1 k ) k ∈ R k\in\mathbb R k ∈ R P P P k = 0 k=0 k = 0 P ′ P' P ′
k 2 = − 2 ( x 1 2 a 4 + y 1 2 b 4 + z 1 2 c 4 ) x 1 2 a 6 + y 1 2 b 6 + z 1 2 c 6 = − 2 S S ′ . k_2=-\frac{2\big(\tfrac{x_1^2}{a^4}+\tfrac{y_1^2}{b^4}+\tfrac{z_1^2}{c^4}\big)}{\tfrac{x_1^2}{a^6}+\tfrac{y_1^2}{b^6}+\tfrac{z_1^2}{c^6}}=-\frac{2S}{S'}. k 2 = − a 6 x 1 2 + b 6 y 1 2 + c 6 z 1 2 2 ( a 4 x 1 2 + b 4 y 1 2 + c 4 z 1 2 ) = − S ′ 2 S . Step 2 — Length of the normal chord P P ′ PP' P P ′
The displacement from P P P P ′ P' P ′ k 2 n k_2\,\mathbf n k 2 n ∣ k 2 ∣ ∣ n ∣ = ∣ k 2 ∣ S |k_2|\,|\mathbf n|=|k_2|\sqrt S ∣ k 2 ∣ ∣ n ∣ = ∣ k 2 ∣ S
P P ′ = 2 S S ′ S = 2 S 3 / 2 S ′ = 2 ( x 1 2 a 4 + y 1 2 b 4 + z 1 2 c 4 ) 3 / 2 x 1 2 a 6 + y 1 2 b 6 + z 1 2 c 6 . \boxed{\;PP'=\frac{2S}{S'}\sqrt S=\frac{2S^{3/2}}{S'}=\frac{2\big(\tfrac{x_1^2}{a^4}+\tfrac{y_1^2}{b^4}+\tfrac{z_1^2}{c^4}\big)^{3/2}}{\tfrac{x_1^2}{a^6}+\tfrac{y_1^2}{b^6}+\tfrac{z_1^2}{c^6}}.\;} P P ′ = S ′ 2 S S = S ′ 2 S 3/2 = a 6 x 1 2 + b 6 y 1 2 + c 6 z 1 2 2 ( a 4 x 1 2 + b 4 y 1 2 + c 4 z 1 2 ) 3/2 . Step 3 — The point G 3 G_3 G 3 x y xy x y P G 3 PG_3 P G 3
G 3 G_3 G 3 z = 0 z=0 z = 0 z z z z 1 + z 1 c 2 k = 0 z_1+\tfrac{z_1}{c^2}k=0 z 1 + c 2 z 1 k = 0 k G 3 = − c 2 k_{G_3}=-c^2 k G 3 = − c 2
P G 3 = ∣ k G 3 ∣ ∣ n ∣ = c 2 S . PG_3=|k_{G_3}|\,|\mathbf n|=c^2\sqrt S. P G 3 = ∣ k G 3 ∣ ∣ n ∣ = c 2 S . Step 4 — Impose P P ′ = 4 P G 3 PP'=4\,PG_3 P P ′ = 4 P G 3
2 S 3 / 2 S ′ = 4 c 2 S ⟹ 2 S S ′ = 4 c 2 ⟹ S = 2 c 2 S ′ . \frac{2S^{3/2}}{S'}=4c^2\sqrt S\ \Longrightarrow\ \frac{2S}{S'}=4c^2\ \Longrightarrow\ S=2c^2 S'. S ′ 2 S 3/2 = 4 c 2 S ⟹ S ′ 2 S = 4 c 2 ⟹ S = 2 c 2 S ′ . Write this out:
x 1 2 a 4 + y 1 2 b 4 + z 1 2 c 4 = 2 c 2 ( x 1 2 a 6 + y 1 2 b 6 + z 1 2 c 6 ) . \frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}+\frac{z_1^2}{c^4}=2c^2\left(\frac{x_1^2}{a^6}+\frac{y_1^2}{b^6}+\frac{z_1^2}{c^6}\right). a 4 x 1 2 + b 4 y 1 2 + c 4 z 1 2 = 2 c 2 ( a 6 x 1 2 + b 6 y 1 2 + c 6 z 1 2 ) . Collect terms:
x 1 2 ( 1 a 4 − 2 c 2 a 6 ) + y 1 2 ( 1 b 4 − 2 c 2 b 6 ) + z 1 2 ( 1 c 4 − 2 c 2 c 6 ) = 0 , x_1^2\!\left(\frac{1}{a^4}-\frac{2c^2}{a^6}\right)+y_1^2\!\left(\frac{1}{b^4}-\frac{2c^2}{b^6}\right)+z_1^2\!\left(\frac{1}{c^4}-\frac{2c^2}{c^6}\right)=0, x 1 2 ( a 4 1 − a 6 2 c 2 ) + y 1 2 ( b 4 1 − b 6 2 c 2 ) + z 1 2 ( c 4 1 − c 6 2 c 2 ) = 0 , x 1 2 a 6 ( a 2 − 2 c 2 ) + y 1 2 b 6 ( b 2 − 2 c 2 ) − z 1 2 c 4 = 0. \frac{x_1^2}{a^6}(a^2-2c^2)+\frac{y_1^2}{b^6}(b^2-2c^2)-\frac{z_1^2}{c^4}=0. a 6 x 1 2 ( a 2 − 2 c 2 ) + b 6 y 1 2 ( b 2 − 2 c 2 ) − c 4 z 1 2 = 0. Multiply by − 1 -1 − 1 P = ( x , y , z ) P=(x,y,z) P = ( x , y , z )
Answer
x 2 a 6 ( 2 c 2 − a 2 ) + y 2 b 6 ( 2 c 2 − b 2 ) + z 2 c 4 = 0. \boxed{\;\frac{x^2}{a^6}(2c^2-a^2)+\frac{y^2}{b^6}(2c^2-b^2)+\frac{z^2}{c^4}=0.\;} a 6 x 2 ( 2 c 2 − a 2 ) + b 6 y 2 ( 2 c 2 − b 2 ) + c 4 z 2 = 0.