To calculate the verties and foci of the conic section: (x9)2 + (y4)2=1

Question
Conic sections
asked 2021-03-07
To calculate the verties and foci of the conic section: \(\displaystyle{\left({x}{9}\right)}{2}\ +\ {\left({y}{4}\right)}{2}={1}\)

Answers (1)

2021-03-08
Step 1 Formula: eccentricity \(\displaystyle{\left({e}\right)}={1}\ -\ {b}{2}{a}{2}\) Step 2 Calculation: Compare this equation with the standard ellipse equation \(\displaystyle{x}{2}{a}{2}\ +\ {y}{2}{b}{2}={1}\) and we get: \(\displaystyle{a}{2}={81}\ \Rightarrow\ {a}={9}{b}{2}={16}\ \Rightarrow\ {b}={4}\) Now, \(\displaystyle{e}={1}\ -\ {b}{2}{a}{2}={1}\ -\ {1681}={659}\) 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 vertices of the conic section and we get: \(\displaystyle{\left(\pm\ {a},\ {0}\right)}\ \text{and}\ {\left({0},\ \pm\ {b}\right)}={\left(\pm\ {9},\ {0}\right)}\ \text{and}\ {\left({0},\ \pm\ {4}\right)}\) Now, \(\displaystyle\text{Foci}\ ={\left(\pm\ {a}{e},\ {0}\right)}={\left(\pm\ {65},\ {0}\right)}\) Thus, the vertices and foci of the conic section are: \(\displaystyle{\left(\pm\ {9},\ {0}\right)}\ \text{and}\ {\left({0},\ \pm\ {4}\right)}\ \text{and}\ {\left(\pm\ {65},\ {0}\right)}.\)
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