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Let P be a point on the parabola, x^{2}=4y. If the distance of P from the centre of the circle, x^{2}+y^{2}+6x+8=0is minimum, then the equation of the tangent to the parabola at P, is :

  • Option 1)

    x+4y-2=0

  • Option 2)

    x-y+3=0

  • Option 3)

    x+y+1=0

  • Option 4)

    x+2y=0

 

Answers (1)

best_answer

as we have learned

Standard equation of parabola -

x^{2}=4ay

- wherein

 

 

General form of a circle -

x^{2}+y^{2}+2gx+2fy+c= 0
 

- wherein

centre = \left ( -g,-f \right )

radius = \sqrt{g^{2}+f^{2}-c}

  x^{2}= 4y

Let the Coordinate of P be

\left ( 2t,t^{2} \right )

\frac{dy}{dx}= \frac{x}{2}= t

 

OP= \sqrt{(2t+3)^{2}+t^{4}}

OP^{2}= t^{4}+(2t+3)^{2}

\frac{dOP^{2}}{dt}=4t^{3}+4(2t+3)=0

t^{3}+2t+3=0

t=-1 satisfies this equation

so point P is (-2,1)

  equation of tangent \frac{y-1}{x+2} = -1\Rightarrow y-1= -x-2

x+y+1=0

 

 

 


Option 1)

x+4y-2=0

Option 2)

x-y+3=0

Option 3)

x+y+1=0

Option 4)

x+2y=0

Posted by

Himanshu

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