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Ellipse question

2013 · Shift 0 · Q36
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Ellipse question

2013 · Shift 0 · Q36

JEE MainMathematicsEllipseMCQ+4 / −1
The equation of the circle passing through the foci of the ellipse x216+y29=1{{{x^2}} \over {16}} + {{{y^2}} \over 9} = 116x2​+9y2​=1, and having centre at (0,3)(0,3)(0,3) is :
  1. A
    x2+y2−6y−7=0{x^2} + {y^2} - 6y - 7 = 0x2+y2−6y−7=0
  2. B
    x2+y2−6y+7=0{x^2} + {y^2} - 6y + 7 = 0x2+y2−6y+7=0
  3. C
    x2+y2−6y−5=0{x^2} + {y^2} - 6y - 5 = 0x2+y2−6y−5=0
  4. D
    x2+y2−6y+5=0{x^2} + {y^2} - 6y + 5 = 0x2+y2−6y+5=0
View written solutionFree

Correct answer: A

  1. Identify the ellipse parameters

The ellipse is

x216+y29=1\frac{x^2}{16}+\frac{y^2}{9}=116x2​+9y2​=1

Comparing with the standard form

x2a2+y2b2=1,\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a2x2​+b2y2​=1,

we get

a2=16,b2=9.a^2=16,\quad b^2=9.a2=16,b2=9.

So,

a=4,b=3.a=4,\quad b=3.a=4,b=3.
  1. Find the foci of the ellipse

For an ellipse with major axis along the xxx-axis,

c2=a2−b2=16−9=7.c^2=a^2-b^2=16-9=7.c2=a2−b2=16−9=7.

Hence,

c=7.c=\sqrt{7}.c=7​.

Therefore, the foci are

(7,0)and(−7,0).(\sqrt{7},0) \quad \text{and} \quad (-\sqrt{7},0).(7​,0)and(−7​,0).
  1. Use the given centre of the circle

The circle has centre (0,3)(0,3)(0,3).

Its equation in standard form is

(x−0)2+(y−3)2=r2.(x-0)^2+(y-3)^2=r^2.(x−0)2+(y−3)2=r2.

We now find the radius using one of the foci, say (7,0)(\sqrt{7},0)(7​,0).

  1. Compute the radius

Distance from (0,3)(0,3)(0,3) to (7,0)(\sqrt{7},0)(7​,0) is

r2=(7−0)2+(0−3)2=7+9=16.r^2=(\sqrt{7}-0)^2+(0-3)^2=7+9=16.r2=(7​−0)2+(0−3)2=7+9=16.

Thus,

r=4.r=4.r=4.
  1. Write the circle equation

So the circle is

x2+(y−3)2=16.x^2+(y-3)^2=16.x2+(y−3)2=16.

Expanding,

x2+y2−6y+9=16x^2+y^2-6y+9=16x2+y2−6y+9=16 x2+y2−6y−7=0.x^2+y^2-6y-7=0.x2+y2−6y−7=0.
  1. Match with the options

This matches:

Option A:

x2+y2−6y−7=0x^2+y^2-6y-7=0x2+y2−6y−7=0

Therefore, the correct answer is A.

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