Paraboloid
We will cover following topics
Paraboloid
- An elliptic paraboloid is represented by the equation x2a2+y2b2=zc, where c is positive.
- A hyperbolic paraboloid is represented by the equation x2a2−y2b2=zc, where c is positive.
- The condition that the plane lx+my+nz=p may touch the paraboloid ax2+by2=2cz is given by: l2a+m2b+2npc=0 and the point of contact is given by (−lcan,−mcbn,−pn).
- The tangent plane to ax2+by2=2cz at the point (α,β,γ) is given by aαx+bβy=c(z+γ).
- The locus of the point of intersection of three mutually perpendicular tangent planes to the paraboloid ax2+by2=2cz is given by: 2z+c(1a+1b)=0.
- The equation of normal at (α,β,γ) is given by x−αaα=y−βbβ=z−γ−c.
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The feet of the normals drawn from a point (f,g,h) to the paraboloid ax2+by2=2cz are given by:
α=f1+ar, β=g1+br, γ=h+cr,
where r is obtained by solving the equation af2(1+ar)2+bg2(1+br)2=2c(h+cr).
Since this equation is of degree 5, this implies that maximum 5 normals can be drawn from a fixed point to a paraboloid.
PYQs
Paraboloid
1) Prove that, in general, three normals can be drawn from a given point to the paraboloid x2+y2=2az, but if the point lies on the surface
27a(x2+y2)+8(a−z)3=0then two of the three normal coincide.
[10M]
2) Find the equations to the generating lines of the paraboloid (x+y+z)(2x+y−z)=6z which pass through the point (1,1,1).
[2018, 13M]
3) Two perpendicular tangent planes to the paraboloid x2+y2=2z intersect in a straight line in the plane x=0. Obtain the curve to which this straight line touches.
[2015, 13M]
4) Show that locus of a point from which three mutually perpendicular tangent lines can be drawn to the paraboloid x2+y2+2z2=0 is x2+y2+4z=1.
[2012, 20M]
5) Show that the plane 3x+4y+7z+52=0 touches the paraboloid 3x2+40z and find the point of contact.
[2010, 20M]
6) Prove that the normals from the point (α,β,γ) to the paraboloid x2a2+y2b2=2z lie on the cone:
αx−α+βx−β+a2−b2x−γ=0[2009, 20M]
7) Show that the feet of the normals from the point P(α,β,γ),β≠0 on the paraboloid x2+y2=4z lie on the sphere 2β(x2+y2+z2)−(α2+β2)y−2β(2+γ)z=0.
[2007, 15M]