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  • If <img alt="y=\mathrm{m}_{1} x+\mathrm{c}_{1} \text { and } y=\mathrm{m}_{2} x+\mathrm{c}_{2}, \mathrm{~m}_{1} \neq \mathrm{m}_{2}" src="https://entrancecorner.oncodecogs.com/gif.latex?y%3D%5

If y=\mathrm{m}_{1} x+\mathrm{c}_{1} \text { and } y=\mathrm{m}_{2} x+\mathrm{c}_{2}, \mathrm{~m}_{1} \neq \mathrm{m}_{2} are two common tangents of circle x^{2}+y^{2}=2 and parabola y^{2}=x, then the value of 8\left|\mathrm{~m}_{1} \mathrm{~m}_{2}\right| is equal to:

Option: 1

3+4\sqrt{2}


Option: 2

-5+6\sqrt{2}


Option: 3

-4+3\sqrt{2}


Option: 4

7+6\sqrt{2}


Answers (1)

best_answer


Equation of the tangent to the parabola can be written as
\mathrm{y=m x+\left(\frac{1}{4 m}\right) \: \: or \: \: y-m x-\frac{1}{4 m}=0}
Now, the perpendicular distance from \mathrm{(0,0)} to the tangent is equal to the radius of the given circle, so
\mathrm{OM=\sqrt{2} \Rightarrow\left|\frac{-\frac{1}{4 m}}{\sqrt{1+m^{2}}}\right|=\sqrt{2}}
\mathrm{\Rightarrow \: 2\left(16 m^{2}\right)\left(1+m^{2}\right)=1\: \: \mathrm{\Rightarrow 32 m^{4}+32 m^{2}-1=0}}
\mathrm{\Rightarrow m^{2}=\frac{-2^{5}+2^{3} \sqrt{16+2}}{2 \times 2^{5}} }        

(Ignoring the negative sign as it is square function)

\mathrm{\text { Now }\left|8 m_{1} m_{2}\right|=\left|8 m^{2}\right| }
\mathrm{=\left|\frac{-2^{5}+2^{3} \sqrt{16+2}}{2^{3}}\right|=3 \sqrt{2}-4 }

Posted by

Anam Khan

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