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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, $x^{2}+y^{2}+6x+8=0$ is 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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