UPSC 2024 Maths Optional Paper 1 Q3c — Step-by-Step Solution

15 marks · Section A

Cone · Analytic Geometry · asked 14× in 13 yrs · Read the full method →

Question

Find the vertex of the cone $4x^2-y^2+2z^2+2xy-3yz+12x-11y+6z+4=0$ .

Vertex = critical point of the defining polynomial; solve $\nabla F=0$ then verify $F=0$ there.

Step 1 — Set up the gradient equations.

F(x,y,z)=4x^2-y^2+2z^2+2xy-3yz+12x-11y+6z+4.

F_x=8x+2y+12=0\;\Rightarrow\;4x+y+6=0. \tag{1}

F_y=-2y+2x-3z-11=0\;\Rightarrow\;2x-2y-3z=11. \tag{2}

F_z=4z-3y+6=0\;\Rightarrow\;-3y+4z=-6. \tag{3}

Step 2 — Solve the system.

From (1): $y=-4x-6$ .

Substitute into (2): $2x-2(-4x-6)-3z=11\;\Rightarrow\;10x-3z=-1.\quad (2')$

Substitute into (3): $-3(-4x-6)+4z=-6\;\Rightarrow\;12x+4z=-24\;\Rightarrow\;3x+z=-6.\quad (3')$

From $(3')$ : $z=-6-3x$ . Substitute into $(2')$ : $10x-3(-6-3x)=-1\;\Rightarrow\;19x=-19\;\Rightarrow\;x=-1$ .

Then $z=-6-3(-1)=-3$ and $y=-4(-1)-6=-2$ .

Step 3 — Verify $F=0$ at the candidate.

F(-1,-2,-3)=4(1)-4+2(9)+2(2)-3(6)+12(-1)-11(-2)+6(-3)+4=4-4+18+4-18-12+22-18+4=0.\;\checkmark

\boxed{\;\text{Vertex: }(-1,-2,-3).\;}