Answer to Question #104111 in Analytic Geometry for Deepak Rana

Question

Find the equation of the normal to the solid 2x
2 −y
2 +8z
2 = 11 at a point where it
intersects the line x−3 = z =
y+1
2
.

Answer 1

Solid: $f(x,y,z)=2x^2 −y^2 +8z^2 - 11=0$

Line: $x−3 = z = \frac{y+1}{2}$

Determine the point of intersection.

$\begin{matrix} 2x^2 −y^2 +8z^2 & = &11 \\ z & = &x-3 \\ z &= &\frac{y+1}{2} \end{matrix}$

$\begin{matrix} 2x^2 −y^2 +8z^2 & = &11 \\ z & = &x-3 \\ y &= &2z-1 \end{matrix}$

$\begin{matrix} 2x^2 −y^2 +8z^2 & = &11 \\ z & = &x-3 \\ y &= &2(x-3)-1 \end{matrix}$

$\begin{matrix} 2x^2 −y^2 +8z^2 & = &11 \\ z & = &x-3 \\ y &= &2x-7 \end{matrix}$

Substitute $z=x-3 \text{ and } y=2x-7$ into the first equation.

$2x^2 - (2x-7)^2 + 8(x-3)^2=11$

Clear the parentheses and simplify.

$3x^2 - 10x + 6 =0$

Using the formula for the discriminant, obtain the solution of this equation.

$x_1= \frac{5}{3} + \frac{\sqrt{7}}{3}, \quad x_2= \frac{5}{3} - \frac{\sqrt{7}}{3}$

Determine the $y \text{ and }z.$

$z_{1} = \frac{5}{3} +\frac{\sqrt{7}}{3} - 3 = -\frac{4}{3} + \frac{\sqrt{7}}{3}, \quad z_{2} = -\frac{4}{3} - \frac{\sqrt{7}}{3} \\ y_{1} = 2( -\frac{4}{3} + \frac{\sqrt{7}}{3}) - 1 = -\frac{11}{3} + \frac{2\sqrt{7}}{3}, \quad y_{2} = -\frac{11}{3} -\frac{2\sqrt{7}}{3}$

Thus, obtain two points of intersection $P_{1} = \left( \frac{5}{3} + \frac{\sqrt{7}}{3}, -\frac{11}{3} + \frac{2\sqrt{7}}{3}, -\frac{4}{3} + \frac{\sqrt{7}}{3}\right) \text{ and } P_{2} = \left( \frac{5}{3} - \frac{\sqrt{7}}{3}, -\frac{11}{3} - \frac{2\sqrt{7}}{3}, -\frac{4}{3} - \frac{\sqrt{7}}{3}\right).$

The equation of normal to solid at the point $P_{1}$ is

$$\frac{x-x_1}{f_{x}(x_1,y_1,z_1)} = \frac{y-y_1}{f_{y}(x_1,y_1,z_1)} = \frac{z-z_1}{f_{z}(x_1,y_1,z_1)}$$

where $f_{x}, f_y, f_z$ are the partial derivatives of $f$ with respect to $x,y,z$ accordingly. Determine the partial derivatives.

$f_x = 4x \\ f_y = -2y\\ f_z = 16z$

Therefore, the equation of the normal to the solid at the first point is

$$\frac{x-\frac{5}{3}-\frac{\sqrt{7}}{3}}{4\cdot(\frac{5}{3}+\frac{\sqrt{7}}{3})}= \frac{y+\frac{11}{3}-\frac{2\sqrt{7}}{3}}{-2\cdot(-\frac{11}{3}+\frac{2\sqrt{7}}{3})} = \frac{z+\frac{4}{3}-\frac{\sqrt{7}}{3}}{16\cdot(-\frac{4}{3}+\frac{\sqrt{7}}{3})}$$

and the equation of the normal to the solid at the second point is

$$\frac{x-\frac{5}{3}+\frac{\sqrt{7}}{3}}{4\cdot(\frac{5}{3}-\frac{\sqrt{7}}{3})}= \frac{y+\frac{11}{3}+\frac{2\sqrt{7}}{3}}{-2\cdot(-\frac{11}{3}-\frac{2\sqrt{7}}{3})} = \frac{z+\frac{4}{3}+\frac{\sqrt{7}}{3}}{16\cdot(-\frac{4}{3}-\frac{\sqrt{7}}{3})}$$