`\color{blue} ✍️` The equation of an ellipse is simplest if the centre of the ellipse is at the origin and the foci are on the x-axis or y-axis.
`\color{blue} ✍️` The two such possible orientations are shown in Fig 11.26.
`\color{blue} ✍️` We will derive the equation for the ellipse shown above in Fig 11.26 (a) with foci on the `x`-axis.
`\color{blue} ✍️` Let `F_1` and `F_2` be the foci and `O` be the midpoint of the line segment `F_1F_2`.
Let `O` be the origin and the line from `O` through `F_2` be the positive x-axis and that through `F_1` as the negative `x`-axis.
Let, the line through `O` perpendicular to the x-axis be the y-axis. Let the coordinates of `F_1` be `(– c, 0)` and `F_2` be `(c, 0)` (Fig 11.27).
Let `P(x, y)` be any point on the ellipse such that the sum of the distances from `P` to the two foci be `2a` so given
`PF_1 + PF_2 = 2a`
Using the distance formula, we have
`sqrt((x+c)^2 + y^2) + sqrt((x-c)^2+y^2)=2a`
i.e `sqrt((x+c)^2 +y^2) = 2a - sqrt((x-c)^2) +y^2`
Squaring both sides, we get
`(x + c)^2 + y^2 = 4a^2 – 4a sqrt((x-c)^2 + y^2)+(x -c)^2 +y^2`
which on simplification gives
`sqrt((x-c)^2 +y^2) = a-c/a x`
Squaring again and simplifying, we get
`(x^2)/(a^2) + (y^2)/(a^2-c^2) = 1`
i.e `(x^2)/(a^2) + (y^2)/(b^2) = 1` ` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ "Since" (c^2 = a^2 – b^2)`
`color(blue)("Hence any point on the ellipse satisfies")`
`color(blue)((x^2)/(a^2) + (y^2)/(b^2) = 1)`.......................................(2)
`\color{blue} ✍️` `color(red)("Conversely,")` let `P (x, y)` satisfy the equation `(2)` with `0 < c < a`. Then
`y^2 = b^2 (1-(x^2)/(a^2))`
Therefore, `PF_1 = sqrt((x+c)^2 +y^2)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =sqrt((x+c)^2+b^2(a^2-x^2)/(a^2))`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =sqrt((x+c)^2+(a^2-c^2) (a^2-x^2)/(a^2))` ` \ \ \ \ \ \ \ \ \ \("since" b^2 = a^2 – c^2)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sqrt((a+(cx)/a)^2) = a+c/a x`
Similarly `PF_2 = a- c/ax`
Hence `PF_1 + PF_2 = a+c/a x + a-c/a x = 2a`........................................(3)
So, any point that satisfies `(x^2)/(a^2) + (y^2)/(b^2) = 1` satisfies the geometric condition and so
P `(x, y)` lies on the ellipse.
Hence from (2) and (3), we proved that the equation of an ellipse with centre of the origin and major axis along the x-axis is
`(x^2)/(a^2) + (y^2)/(b^2) = 1`
Similarly, we can derive the equation of the ellipse in Fig 11.26 (b) as `color(red)((x^2)/(b^2) + (y^2)/(a^2) =1)`
`\color{orange}{"These two equations are known as standard equations of the ellipses."}`
`color{blue} "Key Point :"` The standard equations of ellipses have centre at the origin and the major and minor axis are coordinate axes. However, the study of the ellipses with centre at any other point, and any line through the centre as major and the minor axes passing through the centre and perpendicular to major axis are beyond the scope here.
From the standard equations of the ellipses (Fig11.26),
`color{maroon}{ul"we have the following observations:"}`
`color{maroon}{1.}` `\color{orange}{"Ellipse is symmetric with respect to both the coordinate axes"}` since if `(x, y)` is a point on the ellipse, then `(– x, y), (x, –y)` and `(– x, –y)` are also points on the ellipse.
`color{maroon}{2.}` The foci always lie on the major axis. The major axis can be determined by finding the intercepts on the axes of symmetry. That is, major axis is along the x-axis if the coefficient of `x^2` has the larger denominator and it is along the y-axis if the coefficient of `y^2` has the larger denominator.
`\color{blue} ✍️` The equation of an ellipse is simplest if the centre of the ellipse is at the origin and the foci are on the x-axis or y-axis.
`\color{blue} ✍️` The two such possible orientations are shown in Fig 11.26.
`\color{blue} ✍️` We will derive the equation for the ellipse shown above in Fig 11.26 (a) with foci on the `x`-axis.
`\color{blue} ✍️` Let `F_1` and `F_2` be the foci and `O` be the midpoint of the line segment `F_1F_2`.
Let `O` be the origin and the line from `O` through `F_2` be the positive x-axis and that through `F_1` as the negative `x`-axis.
Let, the line through `O` perpendicular to the x-axis be the y-axis. Let the coordinates of `F_1` be `(– c, 0)` and `F_2` be `(c, 0)` (Fig 11.27).
Let `P(x, y)` be any point on the ellipse such that the sum of the distances from `P` to the two foci be `2a` so given
`PF_1 + PF_2 = 2a`
Using the distance formula, we have
`sqrt((x+c)^2 + y^2) + sqrt((x-c)^2+y^2)=2a`
i.e `sqrt((x+c)^2 +y^2) = 2a - sqrt((x-c)^2) +y^2`
Squaring both sides, we get
`(x + c)^2 + y^2 = 4a^2 – 4a sqrt((x-c)^2 + y^2)+(x -c)^2 +y^2`
which on simplification gives
`sqrt((x-c)^2 +y^2) = a-c/a x`
Squaring again and simplifying, we get
`(x^2)/(a^2) + (y^2)/(a^2-c^2) = 1`
i.e `(x^2)/(a^2) + (y^2)/(b^2) = 1` ` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ "Since" (c^2 = a^2 – b^2)`
`color(blue)("Hence any point on the ellipse satisfies")`
`color(blue)((x^2)/(a^2) + (y^2)/(b^2) = 1)`.......................................(2)
`\color{blue} ✍️` `color(red)("Conversely,")` let `P (x, y)` satisfy the equation `(2)` with `0 < c < a`. Then
`y^2 = b^2 (1-(x^2)/(a^2))`
Therefore, `PF_1 = sqrt((x+c)^2 +y^2)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =sqrt((x+c)^2+b^2(a^2-x^2)/(a^2))`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =sqrt((x+c)^2+(a^2-c^2) (a^2-x^2)/(a^2))` ` \ \ \ \ \ \ \ \ \ \("since" b^2 = a^2 – c^2)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sqrt((a+(cx)/a)^2) = a+c/a x`
Similarly `PF_2 = a- c/ax`
Hence `PF_1 + PF_2 = a+c/a x + a-c/a x = 2a`........................................(3)
So, any point that satisfies `(x^2)/(a^2) + (y^2)/(b^2) = 1` satisfies the geometric condition and so
P `(x, y)` lies on the ellipse.
Hence from (2) and (3), we proved that the equation of an ellipse with centre of the origin and major axis along the x-axis is
`(x^2)/(a^2) + (y^2)/(b^2) = 1`
Similarly, we can derive the equation of the ellipse in Fig 11.26 (b) as `color(red)((x^2)/(b^2) + (y^2)/(a^2) =1)`
`\color{orange}{"These two equations are known as standard equations of the ellipses."}`
`color{blue} "Key Point :"` The standard equations of ellipses have centre at the origin and the major and minor axis are coordinate axes. However, the study of the ellipses with centre at any other point, and any line through the centre as major and the minor axes passing through the centre and perpendicular to major axis are beyond the scope here.
From the standard equations of the ellipses (Fig11.26),
`color{maroon}{ul"we have the following observations:"}`
`color{maroon}{1.}` `\color{orange}{"Ellipse is symmetric with respect to both the coordinate axes"}` since if `(x, y)` is a point on the ellipse, then `(– x, y), (x, –y)` and `(– x, –y)` are also points on the ellipse.
`color{maroon}{2.}` The foci always lie on the major axis. The major axis can be determined by finding the intercepts on the axes of symmetry. That is, major axis is along the x-axis if the coefficient of `x^2` has the larger denominator and it is along the y-axis if the coefficient of `y^2` has the larger denominator.