How Do You Derive the Equation of an Ellipse for a Semi-Elliptical Ceiling?

[raven_claws]

Homework Statement

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1. a hall that is 10 ft. wide has a ceiling that is a semi-ellipse. the ceiling is 10 ft high at the sides and 12 ft high in the center find its equation with the x-axis horizontal and the origin at the center of the ellipse.

Homework Equations

x^2/a^2 + y^2/b^2 = 1

The Attempt at a Solution

The difference between the height at the ceiling and the sides is 12-10 = 2 ft. This is the value of b.

I assume that the vertices of the ellipse are 10 feet apart, the same as the width of the hall. This means that the major axis is 10 ft. Therefore, a = 10/2 =5.

The equation of the ellipse is x^2/25 + y^2/4 = 1

p.s. i think i got the right answer but i need a better solution.

Homework Statement

the point p1(x1,y1) lies outside the circle whose center is at (h,k) and whose radius is r. if T is the length of the tangent from p1 to the circle, prove that t2=(x1-h)^2+y1-k)^2-r^2

Homework Equations

(x-h)^2+(y-k)^2=r^2

The Attempt at a Solution

P (x1,y1) (x-h)^2+(y-k)^2=r^2

i used distance formula

P (x1,y1) to (x-h)^2+(y-k)^2-r^2=0

= (x1-h)^2+y1-k)^2-r^2

 

[KataKoniK]

Now, to prove that T = t2, we can use the Pythagorean theorem. Since T is the length of the tangent from P1 to the circle, it is perpendicular to the radius at that point. Therefore, we can create a right triangle with the tangent as the hypotenuse, the radius as one of the legs, and the distance from P1 to the center of the circle as the other leg. Using the Pythagorean theorem, we can write:

T^2 = r^2 + (x1-h)^2 + (y1-k)^2

Substituting the equation for (x1-h)^2 + (y1-k)^2-r^2 from the distance formula, we get:

T^2 = t2

Therefore, T = t2, proving the given statement.