Find the equation of tangent plane at (alpha, beta, gamma) to the concoid ax2 + by2 + cz2 = 1. Share with your friends Share 5 Manbar Singh answered this The equation of the given concoid is,ax2 + by2 + cz2 = 1 ....1Now, α, β, γ is any point on 1.Equation of the line passing through α, β, γ is,x-αl = y-βm = z-γn = λsay ....2Any point on this line 2 is λl+α, λm+β, λn+γ.This point will lie on 1, if aλl+α2 + bλm+β2 + cλn+γ2 = 1⇒λ2al2 + bm2 + cn2 + 2λaαl + bβm + cγn + aα2 + bβ2 + cγ2 - 1 = 0 ....3Since, α, β, γ is a point on concoid, thenaα2 + bβ2 + cγ2 = 1Now, equation 3 reduces to λ2al2 + bm2 + cn2 + 2λaαl + bβm + cγn = 0⇒λλal2 + bm2 + cn2 + 2aαl + bβm + cγn = 0⇒λ = 0 or λal2 + bm2 + cn2 + 2aαl + bβm + cγn = 0⇒λ = 0 or λ = -2aαl + bβm + cγnal2 + bm2 + cn2 Thus line 2 will be a tangent if both the roots are equal, then-2aαl + bβm + cγnal2 + bm2 + cn2 = 0⇒aαl + bβm + cγn = 0 .......4Eliminating l, m and n from 2 and 4, we get locus of the tangent line as, aαx-α + bβy-β + cγz-γ = 0⇒aαx + bβy+cγz = aα2 + bβ2 + cγ2⇒aαx + bβy+cγz = 1 as, aα2 + bβ2 + cγ2 = 1 .......5The equation aαx + bβy+cγz = 1 represents a tangent plane to concoid ax2 + by2 + cz2 =1. 1 View Full Answer