← 2013 Paper 1
UPSC Maths 2013 Paper 1 Q3a — Solution 20 marks · Section A
Question
Using Lagrange’s multiplier method, find the shortest distance between the line y = 10 − 2 x y=10-2x y = 10 − 2 x x 2 4 + y 2 9 = 1 \dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}=1 4 x 2 + 9 y 2 = 1
Technique
Lagrange multipliers minimising ( linear functional ) 2 (\text{linear functional})^{2} ( linear functional ) 2
Solution
Strategy. The distance from a point ( x , y ) (x,y) ( x , y ) 2 x + y − 10 = 0 2x+y-10=0 2 x + y − 10 = 0 ∣ 2 x + y − 10 ∣ 5 \dfrac{|2x+y-10|}{\sqrt{5}} 5 ∣2 x + y − 10∣ squared distance (no square root needed) over the ellipse.
Step 1 — Set up Lagrange
Minimise
f ( x , y ) = ( 2 x + y − 10 ) 2 subject to g ( x , y ) = x 2 4 + y 2 9 − 1 = 0. f(x,y)=(2x+y-10)^{2}\qquad\text{subject to}\qquad g(x,y)=\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}-1=0. f ( x , y ) = ( 2 x + y − 10 ) 2 subject to g ( x , y ) = 4 x 2 + 9 y 2 − 1 = 0. ∇ f = 2 ( 2 x + y − 10 ) ( 2 , 1 ) , ∇ g = ( x 2 , 2 y 9 ) . \nabla f=2(2x+y-10)\,(2,1),\quad \nabla g=\!\left(\dfrac{x}{2},\;\dfrac{2y}{9}\right). ∇ f = 2 ( 2 x + y − 10 ) ( 2 , 1 ) , ∇ g = ( 2 x , 9 2 y ) . Let k = 2 x + y − 10 k=2x+y-10 k = 2 x + y − 10 ∇ f = λ ∇ g \nabla f=\lambda\nabla g ∇ f = λ ∇ g
4 k = λ x 2 , 2 k = 2 λ y 9 , 4k=\dfrac{\lambda x}{2},\qquad 2k=\dfrac{2\lambda y}{9}, 4 k = 2 λ x , 2 k = 9 2 λ y , which rearrange to
λ x = 8 k , λ y = 9 k . ( ∗ ) \lambda x=8k,\qquad\lambda y=9k.\tag{$\ast$} λ x = 8 k , λ y = 9 k . ( ∗ ) If k = 0 k=0 k = 0 ( x , y ) (x,y) ( x , y ) and on the ellipse — but the line does not meet the ellipse (Step 4 verifies this), so k ≠ 0 k\ne 0 k = 0 λ ≠ 0 \lambda\ne 0 λ = 0 ( ∗ ) (\ast) ( ∗ ) x , y x,y x , y
x = 8 k λ , y = 9 k λ . x=\dfrac{8k}{\lambda},\qquad y=\dfrac{9k}{\lambda}. x = λ 8 k , y = λ 9 k . Step 2 — Use the constraint
x 2 4 + y 2 9 = 64 k 2 4 λ 2 + 81 k 2 9 λ 2 = 16 k 2 + 9 k 2 λ 2 = 25 k 2 λ 2 = 1 ⟹ λ = ± 5 k . \dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}=\dfrac{64k^{2}}{4\lambda^{2}}+\dfrac{81k^{2}}{9\lambda^{2}}=\dfrac{16k^{2}+9k^{2}}{\lambda^{2}}=\dfrac{25k^{2}}{\lambda^{2}}=1\;\Longrightarrow\;\lambda=\pm 5k. 4 x 2 + 9 y 2 = 4 λ 2 64 k 2 + 9 λ 2 81 k 2 = λ 2 16 k 2 + 9 k 2 = λ 2 25 k 2 = 1 ⟹ λ = ± 5 k . Step 3 — Two critical points
Case λ = 5 k \lambda=5k λ = 5 k x = 8 k 5 k = 8 5 , y = 9 k 5 k = 9 5 x=\dfrac{8k}{5k}=\dfrac{8}{5},\;y=\dfrac{9k}{5k}=\dfrac{9}{5} x = 5 k 8 k = 5 8 , y = 5 k 9 k = 5 9
Then k = 2 x + y − 10 = 16 5 + 9 5 − 10 = 25 5 − 10 = 5 − 10 = − 5 k=2x+y-10=\dfrac{16}{5}+\dfrac{9}{5}-10=\dfrac{25}{5}-10=5-10=-5 k = 2 x + y − 10 = 5 16 + 5 9 − 10 = 5 25 − 10 = 5 − 10 = − 5
Distance = ∣ k ∣ 5 = 5 5 = 5 =\dfrac{|k|}{\sqrt 5}=\dfrac{5}{\sqrt 5}=\sqrt 5 = 5 ∣ k ∣ = 5 5 = 5
Case λ = − 5 k \lambda=-5k λ = − 5 k x = − 8 5 , y = − 9 5 x=-\dfrac{8}{5},\;y=-\dfrac{9}{5} x = − 5 8 , y = − 5 9
Then k = − 16 5 − 9 5 − 10 = − 5 − 10 = − 15 k=-\dfrac{16}{5}-\dfrac{9}{5}-10=-5-10=-15 k = − 5 16 − 5 9 − 10 = − 5 − 10 = − 15
Distance = 15 5 = 3 5 =\dfrac{15}{\sqrt 5}=3\sqrt 5 = 5 15 = 3 5
The minimum is 5 \sqrt 5 5 ( 8 5 , 9 5 ) \bigl(\dfrac{8}{5},\dfrac{9}{5}\bigr) ( 5 8 , 5 9 ) 3 5 3\sqrt 5 3 5 ( − 8 5 , − 9 5 ) \bigl(-\dfrac{8}{5},-\dfrac{9}{5}\bigr) ( − 5 8 , − 5 9 )
Step 4 — Verify line and ellipse don’t meet
Substituting y = 10 − 2 x y=10-2x y = 10 − 2 x
x 2 4 + ( 10 − 2 x ) 2 9 = 1 ⟹ 9 x 2 + 4 ( 10 − 2 x ) 2 = 36 ⟹ 25 x 2 − 160 x + 364 = 0. \dfrac{x^{2}}{4}+\dfrac{(10-2x)^{2}}{9}=1\;\Longrightarrow\;9x^{2}+4(10-2x)^{2}=36\;\Longrightarrow\;25x^{2}-160x+364=0. 4 x 2 + 9 ( 10 − 2 x ) 2 = 1 ⟹ 9 x 2 + 4 ( 10 − 2 x ) 2 = 36 ⟹ 25 x 2 − 160 x + 364 = 0. Discriminant = 160 2 − 4 ( 25 ) ( 364 ) = 25600 − 36400 = − 10800 < 0 =160^{2}-4(25)(364)=25600-36400=-10800<0 = 16 0 2 − 4 ( 25 ) ( 364 ) = 25600 − 36400 = − 10800 < 0 5 > 0 \sqrt 5>0 5 > 0
Step 5 — Conclude
Answer
Shortest distance = 5 , at ( 8 5 , 9 5 ) on the ellipse. \boxed{\;\text{Shortest distance}=\sqrt 5,\;\text{at}\;\Bigl(\dfrac{8}{5},\dfrac{9}{5}\Bigr)\;\text{on the ellipse.}\;} Shortest distance = 5 , at ( 5 8 , 5 9 ) on the ellipse.