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Find the (i)length of the axis(ii)coordi...

Find the (i)length of the axis(ii)coordinate of vertices (iii)coordinate of the focii (iv)ecentricity(v)length of latus rectum
`(x^(2))/(9)-(y^(2))/(16)=1`

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To solve the problem step by step, we will analyze the hyperbola given by the equation \(\frac{x^2}{9} - \frac{y^2}{16} = 1\). ### Step 1: Identify the values of \(a\) and \(b\) The standard form of a hyperbola is given by: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] From the equation \(\frac{x^2}{9} - \frac{y^2}{16} = 1\), we can identify: - \(a^2 = 9 \Rightarrow a = 3\) - \(b^2 = 16 \Rightarrow b = 4\) ### Step 2: Find the length of the transverse axis The length of the transverse axis of a hyperbola is given by: \[ \text{Length of transverse axis} = 2a \] Substituting the value of \(a\): \[ \text{Length of transverse axis} = 2 \times 3 = 6 \] ### Step 3: Find the coordinates of the vertices The vertices of the hyperbola are located at \((\pm a, 0)\). Thus: \[ \text{Coordinates of the vertices} = (3, 0) \text{ and } (-3, 0) \] ### Step 4: Find the coordinates of the foci The distance \(c\) from the center to each focus is given by: \[ c = \sqrt{a^2 + b^2} \] Calculating \(c\): \[ c = \sqrt{9 + 16} = \sqrt{25} = 5 \] The coordinates of the foci are located at \((\pm c, 0)\): \[ \text{Coordinates of the foci} = (5, 0) \text{ and } (-5, 0) \] ### Step 5: Find the eccentricity The eccentricity \(e\) of the hyperbola is given by: \[ e = \frac{c}{a} \] Substituting the values of \(c\) and \(a\): \[ e = \frac{5}{3} \] ### Step 6: Find the length of the latus rectum The length of the latus rectum \(L\) is given by: \[ L = \frac{2b^2}{a} \] Calculating \(L\): \[ L = \frac{2 \times 16}{3} = \frac{32}{3} \] ### Summary of Results 1. Length of the transverse axis: \(6\) 2. Coordinates of the vertices: \((3, 0)\) and \((-3, 0)\) 3. Coordinates of the foci: \((5, 0)\) and \((-5, 0)\) 4. Eccentricity: \(\frac{5}{3}\) 5. Length of the latus rectum: \(\frac{32}{3}\)
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