Answer to Question #202769 in Analytic Geometry for tanya

Question

Find the equation of the normal of the paraboloid 3x² +4z² + 4y = 0 at the point (2,−4,1). Also find the point where this line again intersects the paraboloid.

Answer 1

Given paraboloid is f(x,y,z) = 3x²+4y+4z² = 0

Given point is (2,-4,1)

Now gradient g(x,y,z) = d/dx f(x,y,z) i+ d/dy f(x,y,z) j +d/dz f(x,y,z) k

Now g(x,y,z) = 6x i + 4 j + 8z k

g(2,-4,1) = 12 i +4 j + 8 k

The normal line can be parameterized by

$\sigma$(t) = (2 i - 4 j + k) + t(12 i + 4 j +8 k)

$\sigma$(t) = (2+12t)i + (-4+4t) j + (1+8t) k is the equation of normal of paraboloid

It intersect the paraboloid, when f (σ(t)) = 0, meaning that t satisfies.

Now, 3(2+12t)² + 4(-4+4t) + 4(1+8t)² = 0

788 t² + 224 t = 0

t = 0, t = (-56/197)

So, when t = 0, point is (2,-4,1)

When t = (-56/197) , point is ((-278/197),(-1012/197),(-251/197))