Question

Find the equation of ellipse whose foci are at (+- 3,0) and which passes through (4,1)

Answer 1

Answer: The answer is $\dfrac{x^2}{18}+\dfrac{y^2}{9}=1.$

Step-by-step explanation: Given that the foci of an ellipse is given by

$(\pm c,0)=(\pm 3,0).$

And the ellipse passes through the point (4,1). Based on the given information, the equation of the ellipse will be of the form

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1,$

where

$c^2=a^2-b^2\\\\\Rightarrow 3^2=a^2-b^2\\\\\Rightarrow  9=a^2-b^2\\\\\Rightarrow a^2=b^2+9.$

Since the ellipse passes through (4,1), so we have

$\dfrac{4^2}{a^2}+\dfrac{1^2}{b^2}=1\\\\\\\Rightarrow \dfrac{16}{b^2+9}+\dfrac{1}{b^2}=1\\\\\\\Rightarrow 16b^2+b^2+9=b^4+9b^2\\\\\Rightarrow b^4-8b^2-9=0\\\\\Rightarrow b^4-9b^2+b^2-9=0\\\\\Rightarrow (b^2-9)(b^2+1)=0.$

This will give

b² = 9  and  b² = - 1.

Since square of a real number cannot be negative, so b² = 9. Hence,

a² = 9 + 9 = 18.

Thus, the equation of the ellipse is

$\dfrac{x^2}{18}+\dfrac{y^2}{9}=1.$

Answer 2

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