What is the equation of the ellipse with foci at (5, 0), (-5, 0) and vertices (9, 0), (-9, 0)?

Step-by-step explanation:

General equation of ellipse :

Given: Vertices of ellipse : (± 9,0)

Since the vertices are of the form (± a, 0).

i.e. a=9

Thus, the major axis is along x-axis.

Also, foci of the ellipse = (±5,0)

Since, foci is of the form (± c,0), i.e. c=5

Since

Equation of ellipse :

Hence, the required equation :

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

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answer. d.289 hope that