35-(x+7)
= 35-35x-245
= -35x -210
i’m not 100% sure on this because you used square brackets but i think this is the correct answer
Answer:
The equation of tangent plane to the hyperboloid
.
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](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%2B%5Cfrac%7Bz%5E2%7D%7Bc%5E2%7D%3D1)
The equation of tangent plane at the point ![\left(x_0,y_0,z_0\right)](https://tex.z-dn.net/?f=%5Cleft%28x_0%2Cy_0%2Cz_0%5Cright%29)
( Given)
The equation of hyperboloid
![\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D-%5Cfrac%7Bz%5E2%7D%7Bc%5E2%7D%3D1)
F(x,y,z)=![\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}[c^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D-%5Cfrac%7Bz%5E2%7D%5Bc%5E2%7D)
![F_x=\frac{2x}{a^2},F_y=\frac{2y}{b^2},F_z=-\frac{2z}{c^2}](https://tex.z-dn.net/?f=%20F_x%3D%5Cfrac%7B2x%7D%7Ba%5E2%7D%2CF_y%3D%5Cfrac%7B2y%7D%7Bb%5E2%7D%2CF_z%3D-%5Cfrac%7B2z%7D%7Bc%5E2%7D)
![(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)](https://tex.z-dn.net/?f=%20%28F_x%2CF_y%2CF_z%29%28x_0%2Cy_0%2Cz_0%29%3D%5Cleft%28%5Cfrac%7B2x_0%7D%7Ba%5E2%7D%2C%5Cfrac%7B2y_0%7D%7Bb%5E2%7D%2C-%5Cfrac%7B2z_0%7D%7Bc%5E2%7D%5Cright%29)
The equation of tangent plane at point ![\left(x_0,y_0,z_0\right)](https://tex.z-dn.net/?f=%5Cleft%28x_0%2Cy_0%2Cz_0%5Cright%29)
![\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](https://tex.z-dn.net/?f=%5Cfrac%7B2x_0%7D%7Ba%5E2%7D%28x-x_0%29%2B%5Cfrac%7B2y_0%7D%7Bb%5E2%7D%28y-y_0%29-%5Cfarc%7B2z_0%7D%7Bc%5E2%7D%28z-z_0%29%3D0)
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](https://tex.z-dn.net/?f=%5Cfrac%7B2xx_0%7D%7Ba%5E2%7D%2B%5Cfrac%7B2yy_0%7D%7Bb%5E2%7D-%5Cfrac%7B2zz_0%7D%7Bc%5E2%7D-2%5Cleft%28%5Cfrac%7Bx_0%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By_0%5E2%7D%7Bb%5E2%7D-%5Cfrac%7Bz_0%5E2%7D%7Bc%5E2%7D%5Cright%29%3D0)
The equation of tangent plane
![2\left(\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}\right)=2](https://tex.z-dn.net/?f=2%5Cleft%28%5Cfrac%7Bxx_0%7D%7Ba%5E2%7D%2B%5Cfrac%7Byy_0%7D%7Bb%5E2%7D-%5Cfrac%7Bzz_0%7D%7Bc%5E2%7D%5Cright%29%3D2)
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](https://tex.z-dn.net/?f=%5Cfrac%7Bxx_0%7D%7Ba%5E2%7D%2B%5Cfrac%7Byy_0%7D%7Bb%5E2%7D-%5Cfrac%7Bzz_0%7D%7Bc%5E2%7D%3D0)
Now, there is no problem...
180 degrees is the answer