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Question: An ellipse has foci at \({F_1}\,\,\left( {9,20} \right)\) and \({F_2}\,\,\left( {49,55} \right)\) in...

An ellipse has foci at F1  (9,20){F_1}\,\,\left( {9,20} \right) and F2  (49,55){F_2}\,\,\left( {49,55} \right) in the XYXY plane and is tangent to the XX- axis, find the length of its major axis.

Explanation

Solution

Here, the two foci of an ellipse are given and also this ellipse is tangent to the XX- axis. With these data draw a rough figure of the ellipse and then take the image of focus F2{F_2} in XX-axis and we get a straight line whose length is equal to the length of the major axis of the ellipse. And by finding the length of by distance formula we get the required length of its major axis.

Complete step by step solution: Given, the foci of an ellipse are F1  (9,20){F_1}\,\,\left( {9,20} \right) and F2  (49,55){F_2}\,\,\left( {49,55} \right), and ellipse is tangent to XX- axis
We have studied a relation about ellipse that is PF1+PF2=2aP{F_1} + P{F_2} = 2a where aa is the length of the semi major axis.
Since, triangle F2Pq{{F_2}Pq} and triangle F2’Pq{F_2’Pq} are congruent so, the length of sides PF2’{P{F_2’}} and PF2P{F_2}
are equal. So, the above relation can be written as PF2’+PF1=2aP{F_2’} + P{F_1} = 2a which is the length of line segment F1PF2’{{F_1}P{F_2’}}.
Now, find the coordinate of F2’{F_2’} which is the image of F2{F_2} in the abscissa of the coordinate system.
We know that when we take an image of a point in the XX- axis its abscissa of image remains same and ordinate becomes the negative of the ordinate of a given point. So, the coordinate of point F2′F_2' is (49,−55)\left( {49, - 55} \right).
Now, find the length of F1PF2′{F_1}PF_2' using distance formula.
F1F2’=(9−49)2+(20−(−55))2{F_1}{F_2’} = \sqrt {{{\left( {9 - 49} \right)}^2} + {{\left( {20 - \left( { - 55} \right)} \right)}^2}}
F1F2’=(−40)2+(20+55)2{F_1}{F_2’} = \sqrt {{{\left( { - 40} \right)}^2} + {{\left( {20 + 55} \right)}^2}}
F1F2’=1600+(75)2{F_1}{F_2’}= \sqrt {1600 + {{\left( {75} \right)}^2}}
F1F2’=1600+5625{F_1}{F_2’} = \sqrt {1600 + 5625}
F1F2’=7225{F_1}{F_2’} = \sqrt {7225}
F1F2’=85{F_1}{F_2’} = 85
So, the length of line segment F1PF2’{F_1}P{F_2’} is F1F2’=85{F_1}{F_2’} = 85 units.

Thus, the length of the major axis is 8585 units.

Note: Now, in triangle F2Pq{F_2}Pq and triangle F2’Pq{F}_2’Pq
Side pqpq is common in both triangles. ---------(1)
∠F2qP=∠F2’qP=90∘\angle {F_2}qP = \angle {F}_2’qP = {90^ \circ } ----------(2)
In a plane mirror the distance of an object from a mirror is the same as distance between image and mirror.
So, qF2=qF2′q{F_2} = q{F}_2'----------------(3)
By combining all three cases the triangle F2Pq{F_2}Pq and triangle F2’Pq{F}_2’Pq are congruent by side angle side (SAS) property.