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If the common tangents to the purabola, $\mathrm{x^2=4 y}$ and the circle. $\mathrm{x^2+y^2=4}$ interiect at the poimt $\mathrm{P}$ , then the distanice of $\mathrm{P}$ from the origin, is

Option: 1
$2(\sqrt{2}+1)$

Option: 2
$3+2 \sqrt{2}$

Option: 3
$2 \sqrt{(3+2 \sqrt{2})}$

Option: 4
$\sqrt{2}+1$

Answers (1)

best_answer

$\mathrm{\text{Tangent to} \: x^2+y^2=4\: is}$

$\mathrm{y=m x \pm 2 \sqrt{1+m^2}}$

$\mathrm{\text{Given equation of parabola is }x^2=4 y}$

$\mathrm{\therefore \quad x^2=4 m x+8 \sqrt{1+m^2} }$ (From (i))

$\mathrm{ \Rightarrow \quad x^2-4 m x-8 \sqrt{1+m^2}=0 }$

As there is only one intersection point

$\mathrm{ \therefore \text { Discriminant }=0 }$

$\mathrm{ \Rightarrow \quad 16 m^2+4 \cdot 8 \sqrt{1+m^2}=0 \Rightarrow m^2+2 \sqrt{1+m^2}=0 }$

$\mathrm{ \text { or } m^2=-2 \sqrt{1+m^2} }$

$\mathrm{ \Rightarrow m^4=4+4 m^2 \Rightarrow m^4-4 m^2-4=0 }$

$\mathrm{ \Rightarrow m^2=\frac{4 \pm \sqrt{16+16}}{2}=\frac{4 \pm 4 \sqrt{2}}{2} \Rightarrow m^2=2 \pm 2 \sqrt{2} }$

Put in (i), we get
$\mathrm{ y=2 \sqrt{1+2+2 \sqrt{2}}=2 \sqrt{3+2 \sqrt{2}} \quad(\because \text { at } P, x=0) }$

Hence option 3 is correct.

Posted by

manish painkra

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