If the tangent to the parabola  y^{2}=x  at a point  (\alpha,\beta),(\beta>0)  is also a tangent to the ellipse x^{2}+2y^{2}=1 , then  \alpha is equal to :

  • Option 1)

    \sqrt{2}-1

  • Option 2)

    2\sqrt{2}-1

     

  • Option 3)

    2\sqrt{2}+1

  • Option 4)

    \sqrt{2} +1

 

Answers (1)
V Vakul

y^{2}=x

since (\alpha,\beta)   lies on parabola

\\\beta^{2}=\alpha\\\\\:(\alpha,\beta)=(\beta^{2},\beta)\\\\\:tangent\:\:to\:\:parabola\\\\\:y.\beta=\frac{1}{2}\left ( x+\beta^{2} \right )\\\\\:y=\frac{x}{2\beta}+\frac{\beta}{2}

tangent to parabola is also tangent to ellipse

\\\frac{x^{2}}{4}=1.\left ( \frac{1}{4\beta} \right )+\frac{1}{2}\\\\\:\beta^{2}=\sqrt{2}+1,\alpha=\sqrt{2}+1

 


Option 1)

\sqrt{2}-1

Option 2)

2\sqrt{2}-1

 

Option 3)

2\sqrt{2}+1

Option 4)

\sqrt{2} +1

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