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Find the latus rectum, the eccentricity,...

Find the latus rectum, the eccentricity, and the coordinates of the foci, of the ellipses
` 5 x^(2) + 4y^(2) = 1 `

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To solve the problem of finding the latus rectum, eccentricity, and coordinates of the foci of the ellipse given by the equation \(5x^2 + 4y^2 = 1\), we will follow these steps: ### Step 1: Rewrite the equation in standard form The standard form of an ellipse is given by: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] where \(a^2\) and \(b^2\) are the denominators corresponding to the \(x^2\) and \(y^2\) terms respectively. Starting with the given equation: \[ 5x^2 + 4y^2 = 1 \] we can rewrite it as: \[ \frac{x^2}{\frac{1}{5}} + \frac{y^2}{\frac{1}{4}} = 1 \] This implies: \[ a^2 = \frac{1}{5} \quad \text{and} \quad b^2 = \frac{1}{4} \] ### Step 2: Identify \(a\) and \(b\) Calculating \(a\) and \(b\): \[ a = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \quad \text{and} \quad b = \sqrt{\frac{1}{4}} = \frac{1}{2} \] ### Step 3: Determine the latus rectum The formula for the latus rectum \(L\) of an ellipse is given by: \[ L = \frac{2b^2}{a} \] Substituting the values of \(b^2\) and \(a\): \[ L = \frac{2 \cdot \frac{1}{4}}{\frac{1}{\sqrt{5}}} = \frac{\frac{1}{2}}{\frac{1}{\sqrt{5}}} = \frac{1}{2} \cdot \sqrt{5} = \frac{\sqrt{5}}{2} \] ### Step 4: Calculate the eccentricity The eccentricity \(e\) of an ellipse is given by: \[ e = \sqrt{1 - \frac{a^2}{b^2}} \] Substituting the values of \(a^2\) and \(b^2\): \[ e = \sqrt{1 - \frac{\frac{1}{5}}{\frac{1}{4}}} = \sqrt{1 - \frac{4}{5}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \] ### Step 5: Find the coordinates of the foci The coordinates of the foci of an ellipse centered at the origin with a vertical major axis are given by: \[ (0, \pm be) \] Calculating \(be\): \[ be = \frac{1}{2} \cdot \frac{1}{\sqrt{5}} = \frac{1}{2\sqrt{5}} \] Thus, the coordinates of the foci are: \[ (0, \pm \frac{1}{2\sqrt{5}}) \] ### Final Results - Latus Rectum: \(\frac{\sqrt{5}}{2}\) - Eccentricity: \(\frac{1}{\sqrt{5}}\) - Coordinates of the Foci: \((0, \frac{1}{2\sqrt{5}})\) and \((0, -\frac{1}{2\sqrt{5}})\)
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