# 1) Find the parametric equations for the tangent line to the curve of intersection of the...

## Question:

1) Find the parametric equations for the tangent line to the curve of intersection of the paraboloid {eq}\displaystyle z = x^2 + y^2 {/eq} and the ellipsoid {eq}\displaystyle 4x^2 + 2y^2 + 3z^2 = 18 {/eq} at the point {eq}\displaystyle (-1,1,2). {/eq}

2) The plane {eq}\displaystyle y + z = 5 {/eq} intersects the cylinder {eq}\displaystyle x^2 + y^2 = 5 {/eq} in an ellipse. Find parametric equations for the tangent lines to this ellipse at the point {eq}\displaystyle (2, 1, 4). {/eq}

## Tangent Line to a Curve of Intersection

The curve of intersection of two surfaces has the tangent line in the tangent planes to the both surfaces,

therefore the tangent vector is perpendicular to both normal vectors of the given surfaces.

The cross product of two vectors is a vector perpendicular on both vectors, therefore the tangent vector to a curve of intersection is in the direction of the cross product of the normal vectors to the surfaces.

## Answer and Explanation:

1) To find the tangent line to the curve of intersection between the paraboloid {eq}\displaystyle z=f(x,y)=x^2+y^2 {/eq} and the ellipsoid {eq}\displaystyle g(x,y,z)=4x^2 + 2y^2 + 3z^2 = 18 {/eq} at the point {eq}\displaystyle (-1,1,2) {/eq}

we will determine the tangent vector to the curve of intersection.

The tangent vector is perpendicular to both normal vectors to the surfaces, therefore it is parallel to the cross product of the normal vectors.

The normal vector to a surface is the gradient vector of the function describing the surface.

So, the normal vectors are

{eq}\displaystyle \begin{align} \mathbf{n_1}(x,y)&=\left\langle f_x(x,y), f_y(x,y), -1\right\rangle =\langle 2x, 2y, -1\rangle \implies \mathbf{n_1}(-1,1)=\langle -2,2,-1\rangle \\ \mathbf{n_2}(x,y,z)&=\left\langle g_x(x,y,z), g_y(x,y,z), g_z(x,y,z)\right\rangle =\langle 8x, 4y, 6z\rangle \implies \mathbf{n_2}(-1,1, 2)=\langle -8, 4,12\rangle =4\langle -2, 1,3\rangle \\ \mathbf{n_1}(x,y) \times \mathbf{n_2}(x,y,z)&= \begin{array}{|ccc|} \mathbf{i} &\mathbf{j}&\mathbf{k}\\ -2&2&-1\\ -2&1&3 \end{array} =\langle 7,8,2 \rangle.\\ &\boxed{\text{ so, the tangent line is } x=-1+7t, 1+8t, 2+2t, t\in\mathbf{R}}. \end{align} {/eq}

2) To find the tangent line to the curve of intersection between the plane {eq}\displaystyle f(x,y,z)=y+z=5 {/eq} and the cylinder {eq}\displaystyle g(x,y,z)=x^2 + y^2 = 5 {/eq} at the point {eq}\displaystyle (2,1,4) {/eq}

we will determine the tangent vector to the curve of intersection.

The tangent vector is perpendicular to both normal vectors to the surfaces, therefore it is parallel to the cross product of the normal vectors.

So, the normal vectors are

{eq}\displaystyle \begin{align} \mathbf{n_1}(x,y,z)&=\left\langle f_x(x,y), f_y(x,y), f_z(x,y)\right\rangle =\langle 0, 1, 1\rangle \implies \mathbf{n_1}(2,1,4)=\langle 0,1,1\rangle \\ \mathbf{n_2}(x,y,z)&=\left\langle g_x(x,y,z), g_y(x,y,z), g_z(x,y,z)\right\rangle =\langle 2x, 2y,0\rangle \implies \mathbf{n_2}(2,1, 4)=\langle 4, 2,0\rangle =2\langle 2, 1,0\rangle \\ \mathbf{n_1}(x,y) \times \mathbf{n_2}(x,y,z)&= \begin{array}{|ccc|} \mathbf{i} &\mathbf{j}&\mathbf{k}\\ 0&1&1\\ 2&1&0 \end{array} =\langle -1,2, -2 \rangle.\\ &\boxed{\text{ so, the tangent line is } x=2-t, 1+2t, 4-2t, t\in\mathbf{R}}. \end{align} {/eq}

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from NY Regents Exam - Geometry: Tutoring Solution

Chapter 1 / Lesson 11