# To calculate: The vertices and foci of the conic section: x29 + y24=1

Question
Conic sections
To calculate: The vertices and foci of the conic section: $$\displaystyle{x}{29}\ +\ {y}{24}={1}$$

2021-02-01
Step 1 Formula Used: Eccentricity PSK(e)=1\ -\ b2a2 Step 2 Calculation: Compare the given equation with the equation of an ellipse $$\displaystyle{x}{2}{a}{2}\ +\ {y}{2}{b}{2}={1}$$ and we get: $$\displaystyle{a}{2}={9}\ \Rightarrow\ {a}={3}{b}{2}={4}\ \Rightarrow\ {b}={2}$$ Now, $$\displaystyle{e}={1}\ -\ {b}{2}{a}{2}={1}\ -\ {49}={53}$$ Vertices are: $$\displaystyle{\left(\pm\ {a},\ {0}\right)}\ \text{and}\ {\left({0},\ \pm\ {b}\right)}$$ Now, put the values of a and b to get the vertice of the conic section and we get: $$\displaystyle{\left(\pm\ {a},\ {0}\right)}\ \text{and}\ {\left({0},\ \pm\ {b}\right)}={\left(\pm\ {3},\ {0}\right)}\ \text{and}\ {\left({0},\ \pm\ {2}\right)}$$ Now, $$\displaystyle\text{Foci}\ ={\left(\pm\ {a}{e},\ {0}\right)}={\left(\pm\ {5},\ {0}\right)}$$ Thus, the vertices and foci of the conic section are: $$\displaystyle{\left(\pm\ {3},\ {0}\right)}\ \text{and}\ {\left({0},\ \pm\ {2}\right)}\ \text{and}\ {\left(\pm\ {5},\ {0}\right)}.$$

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