Write an equation for the ellipse with vertices (2, 2), (4, 2), (3, -1) and (3,

Question

Write an equation for the ellipse with vertices (2, 2), (4, 2), (3, -1) and (3, 5). Find the foci of the hyperbola (x - 3)^2/4 - (y + 2)^2/1 = 1. What is the standard form equation of a parabola with directrix x = -2 and focus (2, 0)? Eliminate the parameter from the parametric equations x = 1/2t + 6 and y = cos t - sin Determine what kind of conic is represented by the equation r = 1/6 + 4 sin theta. Estimate the following limit using a table. lim_x rightarrow -2 = x^2 - 4/x + 2 Use properties of limits to compute the following limit exactly. lim_x rightarrow 0 = Squareroot x + 7 - Squareroot 7/x Use the limit definition to compute f'(x) if f(x) = 6x^2 - 9.

Explanation / Answer

21 vertices are (2,2) (4,2) (3,-1) (3,5)

standard equation of ellipse is given by

(y-k)^2 / a^2 + (x-h)^2 /b^2 = 1

where h,k is the centre of ellipse

2a = length of major axis whic is 6

a = 6/2 = 3

2b = length of minor axis which is 2

b = 1

centre of the ellipse is midpoint of (2,2) and (4,2) which is (3,2)

hence, equation of ellipse is

(y-2)^2 / 9 + (x-3)^2 / 1 = 1

22) (x-3)^2 / 4 - (y+2)^2/ 1 = 1

standard equation of hyperbola is

(x-h)^2/a^2 - (y-k)^2 / b^2 = 1

comparing the two equations we get

a^2 = 4 , b^2 = 1

c^2 = a^2 + b^2

c^2 = 5

c = sqrt 5

hence foci = ( h+ c , k ) and ( h-c, k )

foci = ( 3+sqrt 5 , -2) and (3-sqrt 5 , -2 )

23) directrix = x=-2

focus = (2,0)

standard equation of parabola is given by

y^2 = 4px

where p is the distance between vertex and directrix and vertex and focus

the distance between directrix and focus is 4 units here

hence , p = 4/2 = 2

therefore, equation of parabola is

y^2 = 8x

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