Answer:
(1) 83.764 million miles
(2) 52.766 million miles
(3) [tex]\frac{x^2}{4900}+\frac{y^2}{2116}=1.[/tex]
Step-by-step explanation:
Let the origin C(0,0) be the center of the elliptical path as shown in the figure, where the location of the sun is at one of the two foci, say f.
The standard equation of the ellipse having the center at the origin is
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\;\cdots(i)[/tex]
where [tex]a[/tex] and [tex]b[/tex] are the semi-axes of the ellipse along the x-axis and y-axis respectively.
Let the points P and A represent the points of perihelion (nearest to the sun) and the aphelion (farthest to the sun) of the closest planet Mercury.
Given that,
CP=46 million miles and
CA=70 million miles.
So, [tex]CP=b[/tex] is the semi-minor axis and [tex]CA=a[/tex] is the semi-major axis.
Let the distances on the axes are in millions of miles. So, the coordinates of the point P and A are [tex]P(0,46)[/tex] and [tex]A(70,0)[/tex] respectively.
(1) From the distance formula, the distance between the perihelion and the aphelion is
[tex]PA=\sqrt{(0-70)^2+(46-0)^2}=83.764[/tex] million miles.
(2) Location of the Sun is at focus, [tex]f[/tex], of the elliptical path.
From the standard relation, the distance of the focus from the center of the ellipse, c, is
[tex]c=ae\;\cdots(ii)[/tex]
where [tex]a[/tex] and [tex]e[/tex] are the semi-major axis and the eccentricity of the ellipse.
The eccentricity of the ellipse is
[tex]e=\sqrt{1-\frac{b^2}{a^2}}[/tex]
[tex]\Rightarrow e=\sqrt{1-\frac{46^2}{70^2}}=0.7538[/tex].
Hence, from the equation (i) the distance of the Sun from the center of the elliptical path of the Mercury is
[tex]c=70\times0.7538=52.766[/tex] million miles.
(3) From the equation (i), the equation of the elliptical orbit of Mercury is
[tex]\frac{x^2}{70^2}+\frac{46^2}{b^2}=1[/tex]
[tex]\Rightarrow \frac{x^2}{4900}+\frac{y^2}{2116}=1.[/tex]
