Answer
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Hint: In this question use the concept that in a rectangular hyperbola the length of transverse axis, length of conjugate axis and length of latus rectum all are equal and it is along the line y = x. Use this to find the coordinates and therefore apply distance formula to get the respective lengths.
Complete Step-by-Step solution:
Given equation of rectangular hyperbola is $xy = {c^2}$...................... (1)
Now as we know that in a rectangular hyperbola the length of transverse axis, length of conjugate axis and length of latus rectum all are equal and it is along y = x................. (2).
So from equation (1) we have,
$ \Rightarrow x.x = {c^2}$
$ \Rightarrow {x^2} = {c^2}$
Now take square root on both sides we have,
$ \Rightarrow x = \sqrt {{c^2}} = \pm c$
Now from equation (2) we have,
$ \Rightarrow y = \pm c$
Therefore (x, y) = (c, c) and (-c, -c)
Now as we know distance between two points ($x_1$, $y_1$) and ($x_2$, $y_2$) is
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Let ($x_1$, $y_1$) = (c, c)
And ($x_2$, $y_2$) = (-c, -c) is
Therefore the distance is
$ \Rightarrow d = \sqrt {{{\left( { - c - c} \right)}^2} + {{\left( { - c - c} \right)}^2}} = \sqrt {4{c^2} + 4{c^2}} = 2c\sqrt 2 $
So the length of transverse axis, length of conjugate axis and length of latus rectum is $2c\sqrt 2 $.
So this is the required answer.
Note: A particular kind of hyperbola in which lengths of transverse and conjugate axis are equal is called a rectangular or an equilateral hyperbola. The eccentricity of the rectangular hyperbola is $\sqrt 2 $. The vertices of a rectangular hyperbola is given as $\left( {c,c} \right){\text{ and }}\left( { - c, - c} \right)$, with foci as $\left( {\sqrt 2 c,\sqrt 2 c} \right){\text{ and }}\left( { - \sqrt 2 c, - \sqrt 2 c} \right)$, the directrices is given as $x + y = \pm c$, for general equation of $xy = {c^2}$. The graphical representation of this hyperbola is shown as
Complete Step-by-Step solution:
Given equation of rectangular hyperbola is $xy = {c^2}$...................... (1)
Now as we know that in a rectangular hyperbola the length of transverse axis, length of conjugate axis and length of latus rectum all are equal and it is along y = x................. (2).
So from equation (1) we have,
$ \Rightarrow x.x = {c^2}$
$ \Rightarrow {x^2} = {c^2}$
Now take square root on both sides we have,
$ \Rightarrow x = \sqrt {{c^2}} = \pm c$
Now from equation (2) we have,
$ \Rightarrow y = \pm c$
Therefore (x, y) = (c, c) and (-c, -c)
Now as we know distance between two points ($x_1$, $y_1$) and ($x_2$, $y_2$) is
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Let ($x_1$, $y_1$) = (c, c)
And ($x_2$, $y_2$) = (-c, -c) is
Therefore the distance is
$ \Rightarrow d = \sqrt {{{\left( { - c - c} \right)}^2} + {{\left( { - c - c} \right)}^2}} = \sqrt {4{c^2} + 4{c^2}} = 2c\sqrt 2 $
So the length of transverse axis, length of conjugate axis and length of latus rectum is $2c\sqrt 2 $.
So this is the required answer.
Note: A particular kind of hyperbola in which lengths of transverse and conjugate axis are equal is called a rectangular or an equilateral hyperbola. The eccentricity of the rectangular hyperbola is $\sqrt 2 $. The vertices of a rectangular hyperbola is given as $\left( {c,c} \right){\text{ and }}\left( { - c, - c} \right)$, with foci as $\left( {\sqrt 2 c,\sqrt 2 c} \right){\text{ and }}\left( { - \sqrt 2 c, - \sqrt 2 c} \right)$, the directrices is given as $x + y = \pm c$, for general equation of $xy = {c^2}$. The graphical representation of this hyperbola is shown as
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