UPSC 2013 Maths Optional Paper 1 Q3a — Step-by-Step Solution

20 marks · Section A

Lagrange's method of multipliers (constrained extrema) · Calculus · asked 8× in 14 yrs · Read the full method →

Question

Using Lagrange’s multiplier method, find the shortest distance between the line $y=10-2x$ and the ellipse $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}=1$ .

Technique

Lagrange multipliers minimising $(\text{linear functional})^{2}$ on the ellipse; the two critical points are antipodal under the gradient direction.

Solution

Strategy. The distance from a point $(x,y)$ to the line $2x+y-10=0$ is $\dfrac{|2x+y-10|}{\sqrt{5}}$ . Minimise the squared distance (no square root needed) over the ellipse.

Step 1 — Set up Lagrange

Minimise

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.

\nabla f=2(2x+y-10)\,(2,1),\quad \nabla g=\!\left(\dfrac{x}{2},\;\dfrac{2y}{9}\right).

Let $k=2x+y-10$ . The Lagrange equations $\nabla f=\lambda\nabla g$ are

4k=\dfrac{\lambda x}{2},\qquad 2k=\dfrac{2\lambda y}{9},

which rearrange to

\lambda x=8k,\qquad\lambda y=9k.\tag{$\ast$}

If $k=0$ then $(x,y)$ lies on the line and on the ellipse — but the line does not meet the ellipse (Step 4 verifies this), so $k\ne 0$ . Hence $\lambda\ne 0$ , and we can solve $(\ast)$ for $x,y$ :

x=\dfrac{8k}{\lambda},\qquad y=\dfrac{9k}{\lambda}.

Step 2 — Use the constraint

\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.

Step 3 — Two critical points

Case $\lambda=5k$ : $x=\dfrac{8k}{5k}=\dfrac{8}{5},\;y=\dfrac{9k}{5k}=\dfrac{9}{5}$ .

Then $k=2x+y-10=\dfrac{16}{5}+\dfrac{9}{5}-10=\dfrac{25}{5}-10=5-10=-5$ .

Distance $=\dfrac{|k|}{\sqrt 5}=\dfrac{5}{\sqrt 5}=\sqrt 5$ .

Case $\lambda=-5k$ : $x=-\dfrac{8}{5},\;y=-\dfrac{9}{5}$ .

Then $k=-\dfrac{16}{5}-\dfrac{9}{5}-10=-5-10=-15$ .

Distance $=\dfrac{15}{\sqrt 5}=3\sqrt 5$ .

The minimum is $\sqrt 5$ at $\bigl(\dfrac{8}{5},\dfrac{9}{5}\bigr)$ ; the maximum is $3\sqrt 5$ at $\bigl(-\dfrac{8}{5},-\dfrac{9}{5}\bigr)$ (the antipodal point on the ellipse — same direction in the gradient sense, opposite physical location).

Step 4 — Verify line and ellipse don’t meet

Substituting $y=10-2x$ into the ellipse equation:

\dfrac{x^{2}}{4}+\dfrac{(10-2x)^{2}}{9}=1\;\Longrightarrow\;9x^{2}+4(10-2x)^{2}=36\;\Longrightarrow\;25x^{2}-160x+364=0.

Discriminant $=160^{2}-4(25)(364)=25600-36400=-10800<0$ . No real intersection ✓ — so the minimum distance is strictly positive, consistent with $\sqrt 5>0$ .

Step 5 — Conclude

Answer

\boxed{\;\text{Shortest distance}=\sqrt 5,\;\text{at}\;\Bigl(\dfrac{8}{5},\dfrac{9}{5}\Bigr)\;\text{on the ellipse.}\;}