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How to solve this problem- - Co-ordinate geometry - JEE Main-9

The area  ( in sq. units ) of the smaller of the two circle that touch the parabola , y^{2}=4x at the point (1,2 ) and the x-axis is :

  • Option 1)

    8\pi(2-\sqrt{2})

  • Option 2)

    4\pi (2-\sqrt{2})

  • Option 3)

    4\pi (3+\sqrt{2})

  • Option 4)

    8\pi (3-2\sqrt{2})

 
Answers (1)
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equation of tangent at (1,2)

=2y=2(x+1)

y=x+1

equation of normal

y=-x+3

let the centre be  (3-r,r)

\\AC^{2}=r^{2}\\\\\:(3-r-1)^{2}+(r-2)^{2}=r^2

\\=(2-r)^{2}+(r-2)^{2}=r^{2}\\\\\:2(r-2)^{2}=r^{2}

\\2r^{2}-8r+8=r^{2}\\\\\:r^{2}-8r+8=0

r=4\pm 2\sqrt{2}

for r=4+ 2\sqrt{2}

3-r \:\:<0

so 

r= 4-2\sqrt{2}

Area= \pir^{2}=8\pi(3-2\sqrt{2})

option 4 is answer


Option 1)

8\pi(2-\sqrt{2})

Option 2)

4\pi (2-\sqrt{2})

Option 3)

4\pi (3+\sqrt{2})

Option 4)

8\pi (3-2\sqrt{2})

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