Question

Equation of circle centered at extremity of positive end of minor axis and passing through the focus on positive x axis of ellipse $\rm \dfrac{x^2}{16} + \dfrac{y^2}{9}=1$ is ??​

Answer 1

Given :  Ellipse x²/16  + y²/9  = 1

Circle centered at extremity of positive end of minor axis and passing through the focus on positive x axis of ellipse

To Find : Equation of Circle

Solution:

x²/16  + y²/9  = 1

=> x²/4² + y²/3² = 1

x²/a² + y²/b² = 1

Horizontal Ellipse

Major axis - at x axis   Ends are ( ± 4 , 0)

Minor axis at y axis   Ends are ( 0 ,  ± 3)

Extremity of positive end of minor axis   = ( 0 , 3)

circle centered at extremity of positive end of minor axis

Hence center of circle = ( 0 , 3)

(x - 0 )² + (y - 3)² = r²

=> x² + (y - 3)² = r²

Foci  ( c , 0)

c² = a² - b²  = 4² - 3² = 9

=> c = ± 3

passing through the focus on positive x axis hence ( 3 , 0)

 x² + (y - 3)² = r²   passes through ( 3, 0)

=> 3² + (0 - 3)² = r²

=> r² = 18

x² + (y - 3)² =  18

Equation of circle centered at extremity of positive end of minor axis and passing through the focus on positive x axis of ellipse x²/16  + y²/9  = 1 is

x² + (y - 3)² =  18

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Answer 2

Answer:

The equation of ellipse is

x² + y²=1

9

The length of semi-major axis is a=3 and the length of semi-minor axis is b=1

The coordinates of point A is (3,0) and the coordinates of point B is (0,1)

The equation of line passing through A,B is x+3y=3

The equation of an auxiliary circle of an ellipse is x²+y² =9

The line AB cuts x²+y²=9 at point M

By solving the above equations

The coordinates of point M are

(− 5/12, 5/9)

The area of triangle AMO is

2 1 ∣0(0− 59)+3( 59−0)− 512 (0−0)∣

= 10

27