# 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) $\frac{2}{17}$

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

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

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

$=2x+y=4$

$Slope=-2 =m_{1}$

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

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

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

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

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

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

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

Option 1)

$\frac{4}{17}$

Option 2)

$\frac{64}{17}$

Option 3)

$\frac{128}{17}$

Option 4)

$\frac{2}{17}$

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