Question

To calculate: The vertices and foci of the conic section: \(\displaystyle{x}{29}\ +\ {y}{24}={1}\)

Answer 1

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: \(\displaystyle{\left(\pm\ {3},\ {0}\right)}\ \text{and}\ {\left({0},\ \pm\ {2}\right)}\ \text{and}\ {\left(\pm\ {5},\ {0}\right)}.\)