The equation of the circle passing through the foci of the ellipse x 2 1 6 + y 2 9 = 1 , and having centre at 0 , 3 is

For x 2 1 6 + y 2 9 = 1 a e = 1 6 - 9 = 7 Foci = ± 7 , 0 Radius = 7 + 9 = 4 ∴ Equation of circle = x - 0 2 + y - 3 2 = 4 2 ⇒ x 2 + y 2 - 6 y - 7 = 0

Home

Mathematics - Ellipse Question with Solution | TestHub

MathematicsEllipsePosition of pointHard2 minPYQ_2013

MathematicsHardstatement

The equation of the circle passing through the foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ , and having centre at $(0, 3)$ is

Question diagram: The equation of the circle passing through the foci of the e

Options:

Answer:

Solution:

For $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$

$a e = \sqrt{16 - 9} = \sqrt{7}$

Foci $= (\pm \sqrt{7}, 0)$

Radius $= \sqrt{7 + 9} = 4$

$∴$ Equation of circle $= {(x - 0)}^{2} + {(y - 3)}^{2} = 4^{2}$

$\Rightarrow x^{2} + y^{2} - 6 y - 7 = 0$

Stream:JEESubject:MathematicsTopic:EllipseSubtopic:Position of point

⏱ 2mℹ️ Source: PYQ_2013

Doubts & Discussion

Loading discussions...