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Question: An ellipse having foci at (3, 3) and (-4, 4) and passing through the origin has eccentricity equal t...

An ellipse having foci at (3, 3) and (-4, 4) and passing through the origin has eccentricity equal to:
A. 37 B. 27 C. 57 D. 35  {\text{A}}{\text{. }}\dfrac{3}{7} \\\ {\text{B}}{\text{. }}\dfrac{2}{7} \\\ {\text{C}}{\text{. }}\dfrac{5}{7} \\\ {\text{D}}{\text{. }}\dfrac{3}{5} \\\

Explanation

Solution

Hint: To find the eccentricity we find distance between foci, we find the sum of distances from foci. Using that we find the eccentricity of the ellipse.

Step-by-step answer:
Let O (0, 0) be the origin and S (3, 3) and S’ (-4, 4) be the foci of the ellipse.

We know, distance between foci in an ellipse = SS’ = 2ae
(Using the formula for distance between 2 points)
D = (x2−x1)2+(y2−y1)2\sqrt {{{\left( {{{\text{x}}_2} - {{\text{x}}_1}} \right)}^2} + {{\left( {{{\text{y}}_2} - {{\text{y}}_1}} \right)}^2}}
⟹SS’ = 2ae = (3+4)2+(3−4)2\sqrt {{{\left( {3 + 4} \right)}^2} + {{\left( {{\text{3}} - 4} \right)}^2}} = 50\sqrt {50}

For any point on ellipse (i.e. the origin) [origin = O (0, 0)]
Sum of distances from foci = SO + S’O = 2a
2a = (3−0)2+(3−0)2\sqrt {{{\left( {3 - 0} \right)}^2} + {{\left( {{\text{3}} - 0} \right)}^2}} +(−4−0)2+(4−0)2\sqrt {{{\left( { - 4 - 0} \right)}^2} + {{\left( {4 - 0} \right)}^2}} = 32+42= 723\sqrt 2 + 4\sqrt 2 = {\text{ 7}}\sqrt 2

Therefore, e = 2ae2a=5072\dfrac{{{\text{2ae}}}}{{{\text{2a}}}} = \dfrac{{\sqrt {50} }}{{7\sqrt 2 }}
= 5272\dfrac{{5\sqrt 2 }}{{7\sqrt 2 }}
= 57\dfrac{5}{7}

Note:The key in solving such types of problems is knowing the formulae of distance between foci (2ae) and the sum of distances from foci (2a). And knowing the formulae in ellipse respectively is a crucial step. e =2ae2a\dfrac{{{\text{2ae}}}}{{{\text{2a}}}}
Distance between two points is given by D =(x2−x1)2+(y2−y1)2\sqrt {{{\left( {{{\text{x}}_2} - {{\text{x}}_1}} \right)}^2} + {{\left( {{{\text{y}}_2} - {{\text{y}}_1}} \right)}^2}} .