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Ksivusya [100]
3 years ago
11

The equation of the tangent plane to the ellipsoid x2/a2 + y2/b2 + z2/c2 = 1 at the point (x0, y0, z0) can be written as xx0 a2

+ yy0 b2 + zz0 c2 = 1 Find the equation of the tangent plane to the hyperboloid x2/a2 + y2/b2 − z2/c2 = 1 at (x0, y0, z0) and express it in a form similar to the one for the ellipsoid.
Mathematics
1 answer:
svetlana [45]3 years ago
7 0

Answer:

The equation of tangent plane to the hyperboloid

\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}=1.

Step-by-step explanation:

Given

The equation of ellipsoid

\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1

The equation of tangent plane at the point \left(x_0,y_0,z_0\right)

\frac{xx_0}{a^2}+\frac{yy_0}{b^2}+\frac{zz_0}{c^2}=1  ( Given)

The equation of hyperboloid

\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1

F(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}[c^2}

F_x=\frac{2x}{a^2},F_y=\frac{2y}{b^2},F_z=-\frac{2z}{c^2}

(F_x,F_y,F_z)(x_0,y_0,z_0)=\left(\frac{2x_0}{a^2},\frac{2y_0}{b^2},-\frac{2z_0}{c^2}\right)

The equation of tangent plane at point \left(x_0,y_0,z_0\right)

\frac{2x_0}{a^2}(x-x_0)+\frac{2y_0}{b^2}(y-y_0)-\farc{2z_0}{c^2}(z-z_0)=0

The equation of tangent plane to the hyperboloid

\frac{2xx_0}{a^2}+\frac{2yy_0}{b^2}-\frac{2zz_0}{c^2}-2\left(\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}-\frac{z_0^2}{c^2}\right)=0

The equation of tangent plane

2\left(\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}\right)=2

Hence, the required equation of tangent plane to the hyperboloid

\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}=0

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