Given:
Major axis on the x-axis and passes through the points (4, 3) and (6, 2).
Since the major axis is on the x-axis, the equation of the ellipse will be the form
\( \dfrac{ x^2}{a^2} – \dfrac{y^2}{b^2} = 1,\)…. (1) [Where ‘a’ is the semi-major axis.]
The ellipse passes through points (4, 3) and (6, 2).
So by putting the values x = 4 and y = 3 in equation (1), we get,
\(\dfrac{ 16}{a^2} + \dfrac{9}{b^2} = 1\) …. (2)
Putting, x = 6 and y = 2 in equation (1), we get,
\(\dfrac{ 36}{a^2} + \dfrac{4}{b^2 }= 1\) …. (3)
From equation (2)
\(\dfrac{ 16}{a^2} = 1 – \dfrac{9}{b^2}\)
\( \dfrac{ 1}{a^2} = \dfrac{1}{16}(1 – \dfrac{9}{b^2})\) …. (4)
Substituting the value of \( \dfrac{ 1}{a^2} \) in equation (3) we get,
b2 = 13
Now substitute the value of b2 in equation (4) we get,
\( \dfrac{ 1}{a^2} = \dfrac{1}{16}(1 – \dfrac{9}{b^2})\)
= \(\dfrac{1} {16}(1 – \dfrac{9}{13})\)
= \(\dfrac{ 1}{16}\dfrac{(13-9)}{13}\)
=\( \dfrac{ 1}{16}(\dfrac{4}{13})\)
= \( \dfrac{ 1}{52}\)
a2 = 52
Equation of ellipse is \( \dfrac{ x^2}{a^2} – \dfrac{y^2}{b^2} = 1,\)
By substituting the values of a2 and b2 in above equation we get,
\(\dfrac{ x^2}{52} + \dfrac{y^2}{13} = 1\)
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