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  • Give answer! Let P be the point on the parabola, y2=8x which is at a minimum distance from the centre C of the circle, x2 + (y+6)2=1. Then the equation of the circle, passing through C and having its centre at P is :

Let P be the point on the parabola, y2=8x which is at a minimum distance from the centre C of the circle, x2 + (y+6)2=1. Then the equation of the circle, passing through C and having its centre at P is :

  • Option 1)

     x2 + y2 − 4x + 8y + 12 = 0

     

  • Option 2)

    x2+y2−x+4y−12=0

     

  • Option 3)

     x^{2}+y^{2}-\frac{x}{4}+2y-24=0

  • Option 4)

    x2 + y2 − 4x + 9y + 18 = 0

     

 

Answers (2)

best_answer

As we learnt in

Parametric coordinates of parabola -

x= at^{2}

y= 2at

- wherein

For the parabola.

y^{2}=4ax

 

 

Equation of a circle -

\left ( x-h \right )^{2}+\left ( y-k \right )^{2}= r^{2}

- wherein

Circle with centre \left ( h,k \right ) and radius r.

 

 Equation of  normal of y^{2}=4ax in

parametric form y=-tx+2at+at^{3}

If it passes through (0,-6)

\Rightarrow -6=2at+at^{3}

Put a=2

we get t= -1

This point is (a, -2a)   \Rightarrow (2, -4)

Radius of circle = distance between  \left ( 0,-6 \right ) and \left ( 2,-4 \right )

\Rightarrow \sqrt{2^{2}+2^{2}}=2\sqrt{2}

Hence equation is

\left ( x-2 \right )^2+\left ( y+4 \right )^2=\left ( 2\sqrt{2} \right )^2

x^{2}+y^{2}-4x+8y+12=0


Option 1)

 x2 + y2 − 4x + 8y + 12 = 0

 

Correct option

Option 2)

x2+y2−x+4y−12=0

 

Incorrect Option

Option 3)

 x^{2}+y^{2}-\frac{x}{4}+2y-24=0

Incorrect Option

Option 4)

x2 + y2 − 4x + 9y + 18 = 0

 

Incorrect Option

Posted by

divya.saini

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