Verified Solution • 2025–26 Session moderate

QJEEM12Ma4b2e: JEE Main 2012 Mathematics - Class 11 JEEM12Ma4b2e Solution: An ellipse is drawn by taking a diameter of thec circle as its…

An ellipse is drawn by taking a diameter of thec circle (x−1)2+y2=1{\left( {x - 1} \right)^2} + {y^2} = 1 as its semi-minor axis and a diameter of the circle x2+(y−2)2=4{x^2} + {\left( {y - 2} \right)^2} = 4 is semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is :

Multiple Choice Options

  • A 4x2+y2=44{x^2} + {y^2} = 4
  • B x2+4y2=8{x^2} + 4{y^2} = 8
  • C 4x2+y2=84{x^2} + {y^2} = 8
  • D x2+4y2=16{x^2} + 4{y^2} = 16

Step-by-Step Verified Solution

Equation of circle is (x−1)2+y2=1{\left( {x - 1} \right)^2} + {y^2} = 1

⇒\Rightarrow radius =1=1 and diameter =2=2

∴\therefore Length of semi-minor axis is 2.2.

Equation of circle is x2+(y−2)2=4=(2)2{x^2} + {\left( {y - 2} \right)^2} = 4 = {\left( 2 \right)^2}

⇒\Rightarrow radius =2=2 and diameter =4=4

∴\therefore Length of semi major axis is 44

We know, equation of ellipse is given by

x2(Major   axis  2)+y2(Minor   axis  2)=1{{{x^2}} \over {\left( {Major\,\,\,axi{s^{\,\,2}}} \right)}} + {{{y^2}} \over {\left( {Minor\,\,\,axi{s^{\,\,2}}} \right)}} = 1

⇒x2(4)2+y2(2)2=1\Rightarrow {{{x^2}} \over {{{\left( 4 \right)}^2}}} + {{{y^2}} \over {{{\left( 2 \right)}^2}}} = 1

⇒x216+y24=1\Rightarrow {{{x^2}} \over {16}} + {{{y^2}} \over 4} = 1

⇒x2+4y2=16\Rightarrow {x^2} + 4{y^2} = 16

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring