Let F1 and F2 be the foci of the ellipse kx^2+y^2=1 where k is greatr than 1 is a constant. Suppose that there is a circle which passes through F1 and F2 and which lies tangent to the ellipse at two points on the x-axis. Compute k.
+26387 Let F1 and F2 be the foci of the ellipse kx^2+y^2=1 where k is greatr than 1 is a constant.
Suppose that there is a circle which passes through F1 and F2 and
which lies tangent to the ellipse at two points on the x-axis.
Compute k.
We rearrange \(kx^2+y^2=1\):
\(\begin{array}{|rcll|} \hline \mathbf{kx^2+y^2} & \mathbf{=} & \mathbf{1} \\ \\ \dfrac{x^2}{\dfrac{1}{k}}+\dfrac{y^2}{1} &=& 1 \\\\ \dfrac{x^2}{ \left(\color{red}{\dfrac{1}{\sqrt{k}}} \right)^2 }+\dfrac{y^2}{{\color{red}1}^2} &=& 1 \quad | \quad \boxed{\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 } \\\\ \mathbf{a} &\mathbf{=} & \mathbf{\dfrac{1}{\sqrt{k}}} \\ \mathbf{b} &\mathbf{=} & \mathbf{1} \\ \hline \end{array} \)
The formula with the focus of this ellipse is \(c^2= b^2-a^2\)
where \(c\) is the distance from the focus to origin.
\(\begin{array}{|rcll|} \hline c^2 &=& b^2-a^2 \quad & | \quad b=1,\ a=\dfrac{1}{\sqrt{k}} \\\\ &=& 1-\dfrac{1}{k} \\\\ \mathbf{c} &\mathbf{=}& \mathbf{\sqrt{1-\dfrac{1}{k}} } \\ \hline \end{array} \)
Circle: \(r = c=a\)
\(\begin{array}{|rcll|} \hline \sqrt{1-\dfrac{1}{k}} &=& \dfrac{1}{\sqrt{k}} \\\\ 1-\dfrac{1}{k} &=& \dfrac{1}{k} \\\\ 1 &=& \dfrac{2}{k} \\\\ \mathbf{k} &\mathbf{=}& \mathbf{2} \\ \hline \end{array}\)
