If the tangents on the ellipse 4x^{2}+y^{2}=8 at the points (1,2) and (a,b) are perpendicular to each other, than a^{2} is equal to :
 

  • Option 1)

    \frac{4}{17}

  • Option 2)

    \frac{64}{17}

  • Option 3)

    \frac{128}{17}

     

  • Option 4)

    \frac{2}{17}

 

Answers (1)

\frac{x^{2}}{(\sqrt{2})^{2}}+\frac{y^{2}}{(2\sqrt{2})^{2}}=1

let (a,b)=\left ( \sqrt{2}\cos \Theta ,2\sqrt{2} \sin \Theta \right )

tangent \; at(1,2)=4x+2y=8

                                 =2x+y=4

Slope=-2 =m_{1}

tangent \; at( \sqrt{2}\cos \Theta ,2\sqrt{2} \sin \Theta)=4 \sqrt{2}\cos \Theta +22\sqrt{2} \sin \Theta=8

                                 Slope=-2\: \cot \Theta =m_{2}

m_{1},m_{2}=-1

-2\times-2\: \cot \Theta =4 \cot \Theta =-1

\Rightarrow \cos\: \theta=\frac{1}{\sqrt{17}}\: or\frac{-1}{\sqrt{17}}

a=\frac{\sqrt{2}}{\sqrt{17}}\: or-\frac{\sqrt{2}}{\sqrt{17}}

a^{2}=\frac{2}{17}


Option 1)

\frac{4}{17}

Option 2)

\frac{64}{17}

Option 3)

\frac{128}{17}

 

Option 4)

\frac{2}{17}

Exams
Articles
Questions