Question

Find parametric equations for the tangent line to the curve of intersection of the paraboloid z = x2 + y2 and the ellipsoid 4x2 + 3y2 + 7z2 = 35 at the point (−1, 1, 2). (Enter your answer as a comma-separated list of equations. Let x, y, and z be in terms of t.)

Answer 1

Answer:

Step-by-step explanation:

A tangent vector to the curve of intersection is given by N1 × N2

Where N1 is normal to the graph of

G(x, y, z) = z= x² + y² at the point P(-1,1,2) and

N2 is normal to the level surface

F(x, y, z)=4x² + 3y²+ 7z²= 35 at the point P(-1,1,2).

Now,

G(x, y, z)=x²+y²-z

Then, taking the grad of G will give a vector perpendicular to the paraboloid

∇G = ∂G/∂x •i + ∂G/∂y •j + ∂G/∂z •k

∇G= 2x•i +2y•j -k

At the point (-1,1,2)

N1=(-2,2,2)

Also to get N2

F(x, y, z)=4x² + 3y²+ 7z² - 35

∇F = ∂F/∂x •i + ∂F/∂y •j + ∂F/∂z •k

∇F = 8x•i + 6y•j + 14z•k

At point (-1,1,2)

N2=(-8,6,28)

Computing N1×N2

Note

i×i=j×j=k×k=0

i×j=k, j×i=-k

j×k=i, k×j=-i

k×i=j, i×k=-j

Then,

a×b= (a•i + b•j + c•k) × (x•i + y•j + z•k)

a×b = a•i×(x•i + y•j + z•k) + b•j×(x•i + y•j + z•k) + c•k×(x•i + y•j + z•k)

a×b= (a•i × x•i)+ (a•i × y•j) + (a•i × z•k) + (b•j × x•i) + (b•j × y•j) + (b•j × z•k) + (c•k × x•i) + (c•k × y•j) + (c•k × z•k)

a×b= 0 + ay•k - az•j - bx•k + 0 + bz•i + cx•j - cy•i + 0

a×b= ay•k - az•j - bx•k + bz•i + cx•j -cy•i

Then, rearranging

a×b= (bz - cy)•i+ (cx - az)•j + (ay-bx)•k

Now let assume that

N1=(-2,2,2) a=-2, b=2 and c=2

N2=(-8,6,28) x=-8, y=6 and z=28

N1×N2=(56-12)•i+(-16+56)•j+(-12+16)•k

N1×N2=44i + 40j + 4k

We compute N1 × N2 = (44,40,4) which is tangent to the curve of intersection at the point P.

The equation of the line is given as

x=r + λt

Where r is the point (-1,1,2)

And λ is the direction (44,40,4)

Hence the tangent line is given by

x = 1 + 44t, y = −1 + 40t, z = 2 + 4t.

Answer 2

The required parametric equation is,

$x=-1+62t\\y=1+64t\\z=2+4t$

Equation of tangent plane:

The formula for the equation of the tangent plane is,

$\left\langle x-x_0,y-y_0,z-z_0 \right\rangle=0$

The given equation is,

$z = x^2 + y^2$

And the ellipsoid $4x^2 + 3y^2 + 7z^2 = 35$ at the point $(-1, 1, 2)$

Let, $F= x^2 + y^2-z$ then,

$\nabla F=\left\langle 2x,2y,-1 \right\rangle$

At (-1,1,2) $\nabla F=\left\langle -2,2,-1 \right\rangle$

And $G=4x^2 + 3y^2 + 7z^2 -35$ then,

$\nabla G=\left\langle 8x,6y,14z \right\rangle$

At (-1,1,2) $\nabla G=\left\langle -8,6,28 \right\rangle$

$n=\left| \begin{matrix}\hat{i} & \hat{j} & \hat{k} \\-2 & 2 & -1 \\ -8& 6& 28\end{matrix} \right|\\=\left\langle 62,64,4 \right\rangle$

Now, applying the above formula we get,

$\left\langle62,64,4 \right\rangle \left\langle x+1,y-1,z-2\right\rangle=0\\62(x+1)+64(y-1)+4(z-2)=0\\31x+32y+2z-5=0$

The parametric equations are,

$x=-1+62t\\y=1+64t\\z=2+4t$

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