Question

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

Answer 1

Hint: Here, the two foci of an ellipse are given and also this ellipse is tangent to the $X$- axis. With these data draw a rough figure of the ellipse and then take the image of focus ${F_2}$ in $X$-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 ${F_1}\,\,\left( {9,20} \right)$ and ${F_2}\,\,\left( {49,55} \right)$, and ellipse is tangent to $X$- axis
We have studied a relation about ellipse that is $P{F_1} + P{F_2} = 2a$ where $a$ is the length of the semi major axis.
Since, triangle ${{F_2}Pq}$ and triangle ${F_2’Pq}$ are congruent so, the length of sides ${P{F_2’}}$ and \[P{F_2}\]
are equal. So, the above relation can be written as $P{F_2’} + P{F_1} = 2a$ which is the length of line segment ${{F_1}P{F_2’}}$.
Now, find the coordinate of ${F_2’}$ which is the image of ${F_2}$ in the abscissa of the coordinate system.
We know that when we take an image of a point in the $X$- axis its abscissa of image remains same and ordinate becomes the negative of the ordinate of a given point. So, the coordinate of point $F_2'$ is $\left( {49, - 55} \right)$.
Now, find the length of ${F_1}PF_2'$ using distance formula.
${F_1}{F_2’} = \sqrt {{{\left( {9 - 49} \right)}^2} + {{\left( {20 - \left( { - 55} \right)} \right)}^2}}$
${F_1}{F_2’} = \sqrt {{{\left( { - 40} \right)}^2} + {{\left( {20 + 55} \right)}^2}} $
${F_1}{F_2’}= \sqrt {1600 + {{\left( {75} \right)}^2}} $
${F_1}{F_2’} = \sqrt {1600 + 5625} $
${F_1}{F_2’} = \sqrt {7225} $
${F_1}{F_2’} = 85$
So, the length of line segment ${F_1}P{F_2’}$ is ${F_1}{F_2’} = 85$ units.

Thus, the length of the major axis is $85$ units.

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