Answer to Question #114948 in Analytic Geometry for ANJU JAYACHANDRAN

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

The equation of the normal:

$\frac{x-x_0}{F'_x(M_0)}=\frac{y-y_0}{F'_y(M_0)}=\frac{z-z_0}{F'_z(M_0)}$

$F'_x=4x$

$F'_y=-2y$

$F'_z=16z$

Then:

$\frac{}{}$ $\frac {x-x_0}{4x_0}=\frac {y-y_0}{-2y_0}=\frac {z-z_0}{16z_0}$

$4x_0=x_1-x_0$

$-2y_0=y_1-y_0$

$16z_0=z_1-z_0$

Where $(x_1,y_1,z_1)$ is the point of intersection of the normal with the given line.

We have:

$x_1=5x_0, y_1=-y_0, z_1=15z_0$

So:

$5x_0-3=15z_0=\frac {-y_0+1}{2}$

$2x_0^2-y_0^2+8z_0^2=11$

$y_0=1-2(5x_0-3), z_0=\frac {5x_0-3}{15}$

$2x_0^2-(1-2(5x_0-3))^2+8(\frac {5x_0-3}{15})^2=11$

$2x_0^2-(1-4(5x_0-3)+4(25x_0^2-30x_0+9))+\frac {8}{225}(25x_0^2-30x_0+9)=11$

We have two cases:

$x_0=1.34, y_0=-6.4, z_0=0.25$

or

$x_0=0.08, y_0=6.2, z_0=-0.17$

Answer:

$\frac {x-1.34}{5.36}=\frac {y+6.4}{12.8}=\frac {z-0.25}{4}$

or

$\frac {x-0.08}{0.32}=\frac {y-6.2}{-12.4}=\frac {z+0.17}{-2.72}$