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  • The distance of the center of the ellipse  <img alt="\mathrm x^2+2 \mathrm y^2-2=0" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%20x%5E2&plus;2%20%5Cmathrm%20y%5E2-2

The distance of the center of the ellipse  \mathrm x^2+2 \mathrm y^2-2=0   to those tangents of the ellipse which are equally inclined to both the axes is

Option: 1

\frac{3}{\sqrt{2}}


Option: 2

\frac{\sqrt{3}}{2}


Option: 3

\frac{\sqrt{2}}{3}


Option: 4

\sqrt{\frac{3}{2}}


Answers (1)

best_answer

Equation of ellipse is   \frac{\mathrm x^2}{2}+\frac{\mathrm y^2}{1}=1  . General tangent to the ellipse of slope \mathrm m is \mathrm y=\mathrm m \mathrm x \pm \sqrt{2 \mathrm m^2+1}.   Since this is equally inclined to axes, so \mathrm m= \pm 1. Thus tangents are  \mathrm y= \pm \: \mathrm x \pm \sqrt{2+1}= \pm\: \mathrm x \pm \sqrt{3}

Distance of any of these tangent these from origin is equal to\sqrt{\frac{3}{2}}.

Hence (4) is the correct answer.

Posted by

Gaurav

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