Find the inclination to the major axis of the diameter of the ellipse, the square of whose length is (1) the arithmetic mean (2) the geometric mean (3) the harmonic mean, between the squares on the major and minor axes.
Answer
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Hint: We use the equation of an ellipse and equation of diameter to find the points of intersection of the diameter and the ellipse. Use the formula of distance between two points to find the length of the diameter and then square it. Square the value of major and minor axes to obtain the terms for three parts to calculate mean. Apply formulas of each of the means and find the value of slope. Equate the value of slope to tangent of the angle and calculate the angle.
* A line through the center of the ellipse is called its diameter. The diameter of the ellipse \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\]is\[y = - \dfrac{{{b^2}}}{{{a^2}m}}x\], where m is the slope of the system of chords\[y = mx + c\].
* Length of a line joining two points \[(x,y)\] and\[(a,b)\] is given by \[L = \sqrt {{{(x - a)}^2} + {{(y - b)}^2}} \]
* If a, b are two observations then:
Arithmetic mean\[ = \dfrac{{a + b}}{2}\]; Geometric mean\[ = \sqrt {ab} \]; Harmonic mean\[ = \dfrac{2}{{\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)}}\]
* Slope of a line \[y = mx + c\]is given by \[m = \tan \theta \]
Complete step-by-step solution:
We have an ellipse \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\].................… (1)
Since the equation of the diameter is \[y = - \dfrac{{{b^2}}}{{{a^2}m}}x\]...............… (2)
Substitute the value of y from equation (2) in equation (1)
\[ \Rightarrow \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{{\left( { - \dfrac{{{b^2}}}{{{a^2}m}}x} \right)}^2}}}{{{b^2}}} = 1\]
\[ \Rightarrow \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{b^4}{x^2}}}{{{a^4}{m^2}{b^2}}} = 1\]
Cancel same terms from numerator and denominator
\[ \Rightarrow \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{b^2}{x^2}}}{{{a^4}{m^2}}} = 1\]
Take LCM in LHS of the equation
\[ \Rightarrow \dfrac{{{a^2}{m^2}{x^2} + {b^2}{x^2}}}{{{a^4}{m^2}}} = 1\]
Cross multiply the values
\[ \Rightarrow \left( {{a^2}{m^2} + {b^2}} \right){x^2} = {a^4}{m^2}\]
Divide both sides by \[{a^2}{m^2} + {b^2}\]
\[ \Rightarrow {x^2} = \dfrac{{{a^4}{m^2}}}{{\left( {{a^2}{m^2} + {b^2}} \right)}}\]
Take square root on both sides of the equation
\[ \Rightarrow \sqrt {{x^2}} = \sqrt {\dfrac{{{a^4}{m^2}}}{{\left( {{a^2}{m^2} + {b^2}} \right)}}} \]
Cancel square power by square root
\[ \Rightarrow x = \pm \dfrac{{{a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }}\]...............… (3)
Substitute the value of x from equation (3) in equation (2) to find the value of y
\[ \Rightarrow y = - \dfrac{{{b^2}}}{{{a^2}m}}\left( { \pm \dfrac{{{a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)\]
Cancel same terms from numerator and denominator
\[ \Rightarrow y = \mp \dfrac{{{b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }}\].................… (4)
So, the points become \[\left( {\dfrac{{{a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }},\dfrac{{ - {b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right);\left( {\dfrac{{ - {a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }},\dfrac{{{b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)\]
These points are points of intersection of the diameter with the ellipse.
We calculate the length of the diameter using a formula of distance between two points.
Length of a line joining two points \[(x,y)\] and\[(a,b)\] is given by \[L = \sqrt {{{(x - a)}^2} + {{(y - b)}^2}} \]
So, the length of diameter (D) is given by
\[D = \sqrt {{{\left( {\dfrac{{{a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }} - \dfrac{{ - {a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)}^2} + {{\left( {\dfrac{{ - {b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }} - \dfrac{{{b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)}^2}} \]
Take LCM in the brackets
\[D = \sqrt {{{\left( {\dfrac{{{a^2}m + {a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)}^2} + {{\left( {\dfrac{{ - {b^2} - {b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)}^2}} \]
\[D = \sqrt {{{\left( {\dfrac{{2{a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)}^2} + {{\left( {\dfrac{{ - 2{b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)}^2}} \]
Square the terms inside the bracket
\[D = \sqrt {\dfrac{{4{a^4}{m^2}}}{{{a^2}{m^2} + {b^2}}} + \dfrac{{4{b^4}}}{{{a^2}{m^2} + {b^2}}}} \]
Take LCM on RHS
\[D = \sqrt {\dfrac{{4{a^4}{m^2} + 4{b^4}}}{{{a^2}{m^2} + {b^2}}}} \]
Square both sides of the equation
\[{D^2} = {\left( {\sqrt {\dfrac{{4{a^4}{m^2} + 4{b^4}}}{{{a^2}{m^2} + {b^2}}}} } \right)^2}\]
Cancel square root by square power
\[{D^2} = \dfrac{{4{a^4}{m^2} + 4{b^4}}}{{{a^2}{m^2} + {b^2}}}\]
Take 4 common from numerator
\[{D^2} = 4\dfrac{{\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}}\] …………….… (5)
So, the square of the length of the diameter is\[4\dfrac{{\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}}\].
We know the major axis and the minor axis of the eclipse are given by 2a and 2b respectively.
Since length of major axis is 2a \[ \Rightarrow {\left( {2a} \right)^2} = 4{a^2}\]
Since length of minor axis is 2b \[ \Rightarrow {\left( {2b} \right)^2} = 4{b^2}\]
Now we calculate the required inclination in each part separately using the observations \[4{a^2},4{b^2}\]
(1) The arithmetic mean:
Arithmetic mean is given by the sum of observations divided by the number of observations.
\[ \Rightarrow \]Mean\[ = \dfrac{{4{a^2} + 4{b^2}}}{2}\]
Cancel 2 from both numerator and denominator
\[ \Rightarrow \]Mean\[ = 2{a^2} + 2{b^2}\]
Now we equate the square of diameter obtained from equation (5) to arithmetic mean
\[ \Rightarrow 4\dfrac{{\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}} = 2{a^2} + 2{b^2}\]
Cancel 2 from both sides of the equation
\[ \Rightarrow \dfrac{{2\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}} = {a^2} + {b^2}\]
Cross multiply values on both sides of the equation
\[ \Rightarrow 2\left( {{a^4}{m^2} + {b^4}} \right) = \left( {{a^2} + {b^2}} \right)\left( {{a^2}{m^2} + {b^2}} \right)\]
Multiply the terms
\[ \Rightarrow 2{a^4}{m^2} + 2{b^4} = {a^4}{m^2} + {a^2}{b^2} + {a^2}{b^2}{m^2} + {b^4}\]
Bring all terms on one side of the equation
\[ \Rightarrow 2{a^4}{m^2} + 2{b^4} - {a^4}{m^2} - {a^2}{b^2} - {a^2}{b^2}{m^2} - {b^4} = 0\]
\[ \Rightarrow {a^4}{m^2} + {b^4} - {a^2}{b^2} - {a^2}{b^2}{m^2} = 0\]
Pair up the terms having common factor between them
\[ \Rightarrow \left( {{a^4}{m^2} - {a^2}{b^2}{m^2}} \right) + \left( {{b^4} - {a^2}{b^2}} \right) = 0\]
Take out common factor from brackets
\[ \Rightarrow {a^2}{m^2}\left( {{a^2} - {b^2}} \right) - {b^2}\left( {{a^2} - {b^2}} \right) = 0\]
Shift negative term to RHS
\[ \Rightarrow {a^2}{m^2}\left( {{a^2} - {b^2}} \right) = {b^2}\left( {{a^2} - {b^2}} \right)\]
Cancel same factor from both sides of the equation
\[ \Rightarrow {a^2}{m^2} = {b^2}\]
Divide both sides by \[{a^2}\]
\[ \Rightarrow {m^2} = \dfrac{{{b^2}}}{{{a^2}}}\]
Take square root on both sides of the equation
\[ \Rightarrow \sqrt {{m^2}} = \sqrt {\dfrac{{{b^2}}}{{{a^2}}}} \]
Cancel square power by square root
\[ \Rightarrow m = \dfrac{b}{a}\]
Since we know slope is given by tangent of the angle of inclination,
\[ \Rightarrow \tan \theta = \dfrac{b}{a}\]
Take inverse tangent on both sides of the equation
\[ \Rightarrow {\tan ^{ - 1}}\left( {\tan \theta } \right) = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)\]
Cancel inverse of the function with the function
\[ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)\]
\[\therefore \]Angle of inclination is \[{\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)\]
(2) The geometric mean:
Since geometric mean of two numbers ‘a’ and ‘b’ \[ = \sqrt {ab} \]
We find Geometric mean of \[4{a^2},4{b^2}\]
\[ \Rightarrow \]Mean\[ = \sqrt {4{a^2} \times 4{b^2}} \]
Cancel square root by square power
\[ \Rightarrow \]Mean\[ = 4ab\]
Now we equate the square of diameter obtained from equation (5) to geometric mean
\[ \Rightarrow 4\dfrac{{\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}} = 4ab\]
Cancel 4 from both sides of the equation
\[ \Rightarrow \dfrac{{\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}} = ab\]
Cross multiply values on both sides of the equation
\[ \Rightarrow \left( {{a^4}{m^2} + {b^4}} \right) = \left( {ab} \right)\left( {{a^2}{m^2} + {b^2}} \right)\]
Multiply the terms
\[ \Rightarrow {a^4}{m^2} + {b^4} = {a^3}b{m^2} + a{b^3}\]
Bring all terms on one side of the equation
\[ \Rightarrow {a^4}{m^2} + {b^4} - {a^3}b{m^2} - a{b^3} = 0\]
Pair up the terms having common factor between them
\[ \Rightarrow \left( {{a^4}{m^2} - {a^3}b{m^2}} \right) + \left( {{b^4} - a{b^3}} \right) = 0\]
Take out common factor from brackets
\[ \Rightarrow {a^3}{m^2}\left( {a - b} \right) - {b^3}\left( {a - b} \right) = 0\]
Shift negative term to RHS
\[ \Rightarrow {a^3}{m^2}\left( {a - b} \right) = {b^3}\left( {a - b} \right)\]
Cancel same factor from both sides of the equation
\[ \Rightarrow {a^3}{m^2} = {b^3}\]
Divide both sides by \[{a^3}\]
\[ \Rightarrow {m^2} = \dfrac{{{b^3}}}{{{a^3}}}\]
Take square root on both sides of the equation
\[ \Rightarrow \sqrt {{m^2}} = \sqrt {{{\left( {\dfrac{b}{a}} \right)}^3}} \]
Cancel square power by square root in LHS
\[ \Rightarrow m = {\left( {\dfrac{b}{a}} \right)^{3/2}}\]
Since we know slope is given by tangent of the angle of inclination,
\[ \Rightarrow \tan \theta = {\left( {\dfrac{b}{a}} \right)^{3/2}}\]
Take inverse tangent on both sides of the equation
\[ \Rightarrow {\tan ^{ - 1}}\left( {\tan \theta } \right) = {\tan ^{ - 1}}{\left( {\dfrac{b}{a}} \right)^{3/2}}\]
Cancel inverse of the function with the function
\[ \Rightarrow \theta = {\tan ^{ - 1}}{\left( {\dfrac{b}{a}} \right)^{3/2}}\]
\[\therefore \]Angle of inclination is \[{\tan ^{ - 1}}{\left( {\dfrac{b}{a}} \right)^{3/2}}\]
(3) The harmonic mean:
Since we know harmonic mean of two numbers ‘a’ and ‘b’ \[ = \dfrac{2}{{\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)}}\]
We find the harmonic mean of \[4{a^2},4{b^2}\]
\[ \Rightarrow \]Mean\[ = \dfrac{2}{{\left( {\dfrac{1}{{4{a^2}}} + \dfrac{1}{{4{b^2}}}} \right)}}\]
Take LCM in the denominator
\[ \Rightarrow \]Mean\[ = \dfrac{2}{{\left( {\dfrac{{4{b^2} + 4{a^2}}}{{16{a^2}{b^2}}}} \right)}}\]
\[ \Rightarrow \]Mean\[ = \dfrac{{2 \times 16{a^2}{b^2}}}{{4\left( {{b^2} + {a^2}} \right)}}\]
Cancel same terms from numerator and denominator
\[ \Rightarrow \]Mean\[ = \dfrac{{2 \times 4{a^2}{b^2}}}{{\left( {{b^2} + {a^2}} \right)}}\]
\[ \Rightarrow \]Mean\[ = \dfrac{{8{a^2}{b^2}}}{{{a^2} + {b^2}}}\]
Now we equate the square of diameter obtained from equation (5) to harmonic mean
\[ \Rightarrow 4\dfrac{{\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}} = \dfrac{{8{a^2}{b^2}}}{{{a^2} + {b^2}}}\]
Cancel 4 from both sides of the equation
\[ \Rightarrow \dfrac{{\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}} = \dfrac{{2{a^2}{b^2}}}{{{a^2} + {b^2}}}\]
Cross multiply values on both sides of the equation
\[ \Rightarrow \left( {{a^4}{m^2} + {b^4}} \right)\left( {{a^2} + {b^2}} \right) = \left( {2{a^2}{b^2}} \right)\left( {{a^2}{m^2} + {b^2}} \right)\]
Multiply the terms
\[ \Rightarrow {a^6}{m^2} + {a^2}{b^4} + {a^4}{b^2}{m^2} + {b^6} = 2{a^4}{b^2}{m^2} + 2{a^2}{b^4}\]
Bring all terms on one side of the equation
\[ \Rightarrow {a^6}{m^2} + {a^2}{b^4} + {a^4}{b^2}{m^2} + {b^6} - 2{a^4}{b^2}{m^2} - 2{a^2}{b^4} = 0\]
\[ \Rightarrow {a^6}{m^2} - {a^2}{b^4} + {b^6} - {a^4}{b^2}{m^2} = 0\]
Pair up the terms having common factor between them
\[ \Rightarrow \left( {{a^6}{m^2} - {a^4}{b^2}{m^2}} \right) + \left( { - {a^2}{b^4} + {b^6}} \right) = 0\]
Take out common factor from brackets
\[ \Rightarrow {a^4}{m^2}\left( {{a^2} - {b^2}} \right) - {b^4}\left( {{a^2} - {b^2}} \right) = 0\]
Shift negative term to RHS
\[ \Rightarrow {a^4}{m^2}\left( {{a^2} - {b^2}} \right) = {b^4}\left( {{a^2} - {b^2}} \right)\]
Cancel same factor from both sides of the equation
\[ \Rightarrow {a^4}{m^2} = {b^4}\]
Divide both sides by \[{a^4}\]
\[ \Rightarrow {m^2} = \dfrac{{{b^4}}}{{{a^4}}}\]
Take square root on both sides of the equation
\[ \Rightarrow \sqrt {{m^2}} = \sqrt {{{\left( {\dfrac{{{b^2}}}{{{a^2}}}} \right)}^2}} \]
Cancel square power by square root
\[ \Rightarrow m = {\left( {\dfrac{b}{a}} \right)^2}\]
Since we know slope is given by tangent of the angle of inclination,
\[ \Rightarrow \tan \theta = {\left( {\dfrac{b}{a}} \right)^2}\]
Take inverse tangent on both sides of the equation
\[ \Rightarrow {\tan ^{ - 1}}\left( {\tan \theta } \right) = {\tan ^{ - 1}}{\left( {\dfrac{b}{a}} \right)^2}\]
Cancel inverse of the function with the function
\[ \Rightarrow \theta = {\tan ^{ - 1}}{\left( {\dfrac{b}{a}} \right)^2}\]
\[\therefore \]Angle of inclination is \[{\tan ^{ - 1}}{\left( {\dfrac{b}{a}} \right)^2}\]
Note: Students many times make mistakes in writing the angle as tangent value, keep in mind the tangent value is the value of the slope, the angle of which we take tan is the angle of inclination. Also, many students make mistakes while shifting values from one side of the equation to another, keep in mind we always change sign when shifting values.
* A line through the center of the ellipse is called its diameter. The diameter of the ellipse \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\]is\[y = - \dfrac{{{b^2}}}{{{a^2}m}}x\], where m is the slope of the system of chords\[y = mx + c\].
* Length of a line joining two points \[(x,y)\] and\[(a,b)\] is given by \[L = \sqrt {{{(x - a)}^2} + {{(y - b)}^2}} \]
* If a, b are two observations then:
Arithmetic mean\[ = \dfrac{{a + b}}{2}\]; Geometric mean\[ = \sqrt {ab} \]; Harmonic mean\[ = \dfrac{2}{{\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)}}\]
* Slope of a line \[y = mx + c\]is given by \[m = \tan \theta \]
Complete step-by-step solution:
We have an ellipse \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\].................… (1)
Since the equation of the diameter is \[y = - \dfrac{{{b^2}}}{{{a^2}m}}x\]...............… (2)
Substitute the value of y from equation (2) in equation (1)
\[ \Rightarrow \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{{\left( { - \dfrac{{{b^2}}}{{{a^2}m}}x} \right)}^2}}}{{{b^2}}} = 1\]
\[ \Rightarrow \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{b^4}{x^2}}}{{{a^4}{m^2}{b^2}}} = 1\]
Cancel same terms from numerator and denominator
\[ \Rightarrow \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{b^2}{x^2}}}{{{a^4}{m^2}}} = 1\]
Take LCM in LHS of the equation
\[ \Rightarrow \dfrac{{{a^2}{m^2}{x^2} + {b^2}{x^2}}}{{{a^4}{m^2}}} = 1\]
Cross multiply the values
\[ \Rightarrow \left( {{a^2}{m^2} + {b^2}} \right){x^2} = {a^4}{m^2}\]
Divide both sides by \[{a^2}{m^2} + {b^2}\]
\[ \Rightarrow {x^2} = \dfrac{{{a^4}{m^2}}}{{\left( {{a^2}{m^2} + {b^2}} \right)}}\]
Take square root on both sides of the equation
\[ \Rightarrow \sqrt {{x^2}} = \sqrt {\dfrac{{{a^4}{m^2}}}{{\left( {{a^2}{m^2} + {b^2}} \right)}}} \]
Cancel square power by square root
\[ \Rightarrow x = \pm \dfrac{{{a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }}\]...............… (3)
Substitute the value of x from equation (3) in equation (2) to find the value of y
\[ \Rightarrow y = - \dfrac{{{b^2}}}{{{a^2}m}}\left( { \pm \dfrac{{{a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)\]
Cancel same terms from numerator and denominator
\[ \Rightarrow y = \mp \dfrac{{{b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }}\].................… (4)
So, the points become \[\left( {\dfrac{{{a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }},\dfrac{{ - {b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right);\left( {\dfrac{{ - {a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }},\dfrac{{{b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)\]
These points are points of intersection of the diameter with the ellipse.
We calculate the length of the diameter using a formula of distance between two points.
Length of a line joining two points \[(x,y)\] and\[(a,b)\] is given by \[L = \sqrt {{{(x - a)}^2} + {{(y - b)}^2}} \]
So, the length of diameter (D) is given by
\[D = \sqrt {{{\left( {\dfrac{{{a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }} - \dfrac{{ - {a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)}^2} + {{\left( {\dfrac{{ - {b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }} - \dfrac{{{b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)}^2}} \]
Take LCM in the brackets
\[D = \sqrt {{{\left( {\dfrac{{{a^2}m + {a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)}^2} + {{\left( {\dfrac{{ - {b^2} - {b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)}^2}} \]
\[D = \sqrt {{{\left( {\dfrac{{2{a^2}m}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)}^2} + {{\left( {\dfrac{{ - 2{b^2}}}{{\sqrt {{a^2}{m^2} + {b^2}} }}} \right)}^2}} \]
Square the terms inside the bracket
\[D = \sqrt {\dfrac{{4{a^4}{m^2}}}{{{a^2}{m^2} + {b^2}}} + \dfrac{{4{b^4}}}{{{a^2}{m^2} + {b^2}}}} \]
Take LCM on RHS
\[D = \sqrt {\dfrac{{4{a^4}{m^2} + 4{b^4}}}{{{a^2}{m^2} + {b^2}}}} \]
Square both sides of the equation
\[{D^2} = {\left( {\sqrt {\dfrac{{4{a^4}{m^2} + 4{b^4}}}{{{a^2}{m^2} + {b^2}}}} } \right)^2}\]
Cancel square root by square power
\[{D^2} = \dfrac{{4{a^4}{m^2} + 4{b^4}}}{{{a^2}{m^2} + {b^2}}}\]
Take 4 common from numerator
\[{D^2} = 4\dfrac{{\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}}\] …………….… (5)
So, the square of the length of the diameter is\[4\dfrac{{\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}}\].
We know the major axis and the minor axis of the eclipse are given by 2a and 2b respectively.
Since length of major axis is 2a \[ \Rightarrow {\left( {2a} \right)^2} = 4{a^2}\]
Since length of minor axis is 2b \[ \Rightarrow {\left( {2b} \right)^2} = 4{b^2}\]
Now we calculate the required inclination in each part separately using the observations \[4{a^2},4{b^2}\]
(1) The arithmetic mean:
Arithmetic mean is given by the sum of observations divided by the number of observations.
\[ \Rightarrow \]Mean\[ = \dfrac{{4{a^2} + 4{b^2}}}{2}\]
Cancel 2 from both numerator and denominator
\[ \Rightarrow \]Mean\[ = 2{a^2} + 2{b^2}\]
Now we equate the square of diameter obtained from equation (5) to arithmetic mean
\[ \Rightarrow 4\dfrac{{\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}} = 2{a^2} + 2{b^2}\]
Cancel 2 from both sides of the equation
\[ \Rightarrow \dfrac{{2\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}} = {a^2} + {b^2}\]
Cross multiply values on both sides of the equation
\[ \Rightarrow 2\left( {{a^4}{m^2} + {b^4}} \right) = \left( {{a^2} + {b^2}} \right)\left( {{a^2}{m^2} + {b^2}} \right)\]
Multiply the terms
\[ \Rightarrow 2{a^4}{m^2} + 2{b^4} = {a^4}{m^2} + {a^2}{b^2} + {a^2}{b^2}{m^2} + {b^4}\]
Bring all terms on one side of the equation
\[ \Rightarrow 2{a^4}{m^2} + 2{b^4} - {a^4}{m^2} - {a^2}{b^2} - {a^2}{b^2}{m^2} - {b^4} = 0\]
\[ \Rightarrow {a^4}{m^2} + {b^4} - {a^2}{b^2} - {a^2}{b^2}{m^2} = 0\]
Pair up the terms having common factor between them
\[ \Rightarrow \left( {{a^4}{m^2} - {a^2}{b^2}{m^2}} \right) + \left( {{b^4} - {a^2}{b^2}} \right) = 0\]
Take out common factor from brackets
\[ \Rightarrow {a^2}{m^2}\left( {{a^2} - {b^2}} \right) - {b^2}\left( {{a^2} - {b^2}} \right) = 0\]
Shift negative term to RHS
\[ \Rightarrow {a^2}{m^2}\left( {{a^2} - {b^2}} \right) = {b^2}\left( {{a^2} - {b^2}} \right)\]
Cancel same factor from both sides of the equation
\[ \Rightarrow {a^2}{m^2} = {b^2}\]
Divide both sides by \[{a^2}\]
\[ \Rightarrow {m^2} = \dfrac{{{b^2}}}{{{a^2}}}\]
Take square root on both sides of the equation
\[ \Rightarrow \sqrt {{m^2}} = \sqrt {\dfrac{{{b^2}}}{{{a^2}}}} \]
Cancel square power by square root
\[ \Rightarrow m = \dfrac{b}{a}\]
Since we know slope is given by tangent of the angle of inclination,
\[ \Rightarrow \tan \theta = \dfrac{b}{a}\]
Take inverse tangent on both sides of the equation
\[ \Rightarrow {\tan ^{ - 1}}\left( {\tan \theta } \right) = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)\]
Cancel inverse of the function with the function
\[ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)\]
\[\therefore \]Angle of inclination is \[{\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)\]
(2) The geometric mean:
Since geometric mean of two numbers ‘a’ and ‘b’ \[ = \sqrt {ab} \]
We find Geometric mean of \[4{a^2},4{b^2}\]
\[ \Rightarrow \]Mean\[ = \sqrt {4{a^2} \times 4{b^2}} \]
Cancel square root by square power
\[ \Rightarrow \]Mean\[ = 4ab\]
Now we equate the square of diameter obtained from equation (5) to geometric mean
\[ \Rightarrow 4\dfrac{{\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}} = 4ab\]
Cancel 4 from both sides of the equation
\[ \Rightarrow \dfrac{{\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}} = ab\]
Cross multiply values on both sides of the equation
\[ \Rightarrow \left( {{a^4}{m^2} + {b^4}} \right) = \left( {ab} \right)\left( {{a^2}{m^2} + {b^2}} \right)\]
Multiply the terms
\[ \Rightarrow {a^4}{m^2} + {b^4} = {a^3}b{m^2} + a{b^3}\]
Bring all terms on one side of the equation
\[ \Rightarrow {a^4}{m^2} + {b^4} - {a^3}b{m^2} - a{b^3} = 0\]
Pair up the terms having common factor between them
\[ \Rightarrow \left( {{a^4}{m^2} - {a^3}b{m^2}} \right) + \left( {{b^4} - a{b^3}} \right) = 0\]
Take out common factor from brackets
\[ \Rightarrow {a^3}{m^2}\left( {a - b} \right) - {b^3}\left( {a - b} \right) = 0\]
Shift negative term to RHS
\[ \Rightarrow {a^3}{m^2}\left( {a - b} \right) = {b^3}\left( {a - b} \right)\]
Cancel same factor from both sides of the equation
\[ \Rightarrow {a^3}{m^2} = {b^3}\]
Divide both sides by \[{a^3}\]
\[ \Rightarrow {m^2} = \dfrac{{{b^3}}}{{{a^3}}}\]
Take square root on both sides of the equation
\[ \Rightarrow \sqrt {{m^2}} = \sqrt {{{\left( {\dfrac{b}{a}} \right)}^3}} \]
Cancel square power by square root in LHS
\[ \Rightarrow m = {\left( {\dfrac{b}{a}} \right)^{3/2}}\]
Since we know slope is given by tangent of the angle of inclination,
\[ \Rightarrow \tan \theta = {\left( {\dfrac{b}{a}} \right)^{3/2}}\]
Take inverse tangent on both sides of the equation
\[ \Rightarrow {\tan ^{ - 1}}\left( {\tan \theta } \right) = {\tan ^{ - 1}}{\left( {\dfrac{b}{a}} \right)^{3/2}}\]
Cancel inverse of the function with the function
\[ \Rightarrow \theta = {\tan ^{ - 1}}{\left( {\dfrac{b}{a}} \right)^{3/2}}\]
\[\therefore \]Angle of inclination is \[{\tan ^{ - 1}}{\left( {\dfrac{b}{a}} \right)^{3/2}}\]
(3) The harmonic mean:
Since we know harmonic mean of two numbers ‘a’ and ‘b’ \[ = \dfrac{2}{{\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)}}\]
We find the harmonic mean of \[4{a^2},4{b^2}\]
\[ \Rightarrow \]Mean\[ = \dfrac{2}{{\left( {\dfrac{1}{{4{a^2}}} + \dfrac{1}{{4{b^2}}}} \right)}}\]
Take LCM in the denominator
\[ \Rightarrow \]Mean\[ = \dfrac{2}{{\left( {\dfrac{{4{b^2} + 4{a^2}}}{{16{a^2}{b^2}}}} \right)}}\]
\[ \Rightarrow \]Mean\[ = \dfrac{{2 \times 16{a^2}{b^2}}}{{4\left( {{b^2} + {a^2}} \right)}}\]
Cancel same terms from numerator and denominator
\[ \Rightarrow \]Mean\[ = \dfrac{{2 \times 4{a^2}{b^2}}}{{\left( {{b^2} + {a^2}} \right)}}\]
\[ \Rightarrow \]Mean\[ = \dfrac{{8{a^2}{b^2}}}{{{a^2} + {b^2}}}\]
Now we equate the square of diameter obtained from equation (5) to harmonic mean
\[ \Rightarrow 4\dfrac{{\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}} = \dfrac{{8{a^2}{b^2}}}{{{a^2} + {b^2}}}\]
Cancel 4 from both sides of the equation
\[ \Rightarrow \dfrac{{\left( {{a^4}{m^2} + {b^4}} \right)}}{{{a^2}{m^2} + {b^2}}} = \dfrac{{2{a^2}{b^2}}}{{{a^2} + {b^2}}}\]
Cross multiply values on both sides of the equation
\[ \Rightarrow \left( {{a^4}{m^2} + {b^4}} \right)\left( {{a^2} + {b^2}} \right) = \left( {2{a^2}{b^2}} \right)\left( {{a^2}{m^2} + {b^2}} \right)\]
Multiply the terms
\[ \Rightarrow {a^6}{m^2} + {a^2}{b^4} + {a^4}{b^2}{m^2} + {b^6} = 2{a^4}{b^2}{m^2} + 2{a^2}{b^4}\]
Bring all terms on one side of the equation
\[ \Rightarrow {a^6}{m^2} + {a^2}{b^4} + {a^4}{b^2}{m^2} + {b^6} - 2{a^4}{b^2}{m^2} - 2{a^2}{b^4} = 0\]
\[ \Rightarrow {a^6}{m^2} - {a^2}{b^4} + {b^6} - {a^4}{b^2}{m^2} = 0\]
Pair up the terms having common factor between them
\[ \Rightarrow \left( {{a^6}{m^2} - {a^4}{b^2}{m^2}} \right) + \left( { - {a^2}{b^4} + {b^6}} \right) = 0\]
Take out common factor from brackets
\[ \Rightarrow {a^4}{m^2}\left( {{a^2} - {b^2}} \right) - {b^4}\left( {{a^2} - {b^2}} \right) = 0\]
Shift negative term to RHS
\[ \Rightarrow {a^4}{m^2}\left( {{a^2} - {b^2}} \right) = {b^4}\left( {{a^2} - {b^2}} \right)\]
Cancel same factor from both sides of the equation
\[ \Rightarrow {a^4}{m^2} = {b^4}\]
Divide both sides by \[{a^4}\]
\[ \Rightarrow {m^2} = \dfrac{{{b^4}}}{{{a^4}}}\]
Take square root on both sides of the equation
\[ \Rightarrow \sqrt {{m^2}} = \sqrt {{{\left( {\dfrac{{{b^2}}}{{{a^2}}}} \right)}^2}} \]
Cancel square power by square root
\[ \Rightarrow m = {\left( {\dfrac{b}{a}} \right)^2}\]
Since we know slope is given by tangent of the angle of inclination,
\[ \Rightarrow \tan \theta = {\left( {\dfrac{b}{a}} \right)^2}\]
Take inverse tangent on both sides of the equation
\[ \Rightarrow {\tan ^{ - 1}}\left( {\tan \theta } \right) = {\tan ^{ - 1}}{\left( {\dfrac{b}{a}} \right)^2}\]
Cancel inverse of the function with the function
\[ \Rightarrow \theta = {\tan ^{ - 1}}{\left( {\dfrac{b}{a}} \right)^2}\]
\[\therefore \]Angle of inclination is \[{\tan ^{ - 1}}{\left( {\dfrac{b}{a}} \right)^2}\]
Note: Students many times make mistakes in writing the angle as tangent value, keep in mind the tangent value is the value of the slope, the angle of which we take tan is the angle of inclination. Also, many students make mistakes while shifting values from one side of the equation to another, keep in mind we always change sign when shifting values.
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