Find the equation for the ellipse that satisfies the given conditions Major axis on the x-axis and passes through the points (4,3) and (6,2).

Asked by Sakshi | 1 year ago |  60

##### Solution :-

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,

$$\dfrac{ x^2}{52} + \dfrac{y^2}{13} = 1$$

Answered by Aaryan | 1 year ago

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