UPSC 2017 Maths Optional Paper 1 Q1d — Step-by-Step Solution

10 marks · Section A

Paraboloid (elliptic and hyperbolic) · Analytic Geometry · asked 6× in 13 yrs · Read the full method →

Question

Find the equation of the tangent plane at point $(1,1,1)$ to the conicoid $3x^2-y^2=2z$ .

Tangent plane $=\nabla F\cdot(\mathbf r-\mathbf r_0)=0$ with $F=3x^2-y^2-2z$ .

Write $F(x,y,z)=3x^2-y^2-2z=0$ . At $(1,1,1)$ : $3(1)-1-2(1)=0$ ✓, so the point is on the conicoid.

The tangent plane at $(x_0,y_0,z_0)$ has normal $\nabla F$ :

F_x=6x,\qquad F_y=-2y,\qquad F_z=-2.

At $(1,1,1)$ :

\nabla F=(6,\,-2,\,-2).

\nabla F\cdot\big((x,y,z)-(1,1,1)\big)=0:\quad 6(x-1)-2(y-1)-2(z-1)=0.

Expand: $6x-6-2y+2-2z+2=0\Rightarrow 6x-2y-2z-2=0$ . Divide by $2$ :

\boxed{\;3x-y-z=1.\;}