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.

The equation of tangent plane to the hyperboloid

.

Step-by-step explanation:

Given

The equation of ellipsoid

The equation of tangent plane at the point

 ( Given)

The equation of hyperboloid

F(x,y,z)=

The equation of tangent plane at point

The equation of tangent plane to the hyperboloid

The equation of tangent plane

Hence, the required equation of tangent plane to the hyperboloid

The equation of the tangent plane to the ellipsoid at the given point is

The equation of the tangent plane to the ellipsoid is given as:

The point on the ellipsoid is given as:

(x0, y0, z0)

The equation of the tangent plane to the ellipsoid at the given point can be written as:

Given that the equation of the tangent plane to the hyperboloid is

The equations at the tangents of the ellipsoid and the hyperboloid take the same form.

So, the equation of the tangent plane to the ellipsoid at the given point is

Read more about tangent planes at:

link

the 1st digit is in the 100s n its 1

step-by-step explanation:

3,589÷18 = 199.389

so the first digit of the quotient is in the 100s n its 1

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.