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

Step 1

Formula Used: Eccentricity $(e)=1-b^2{a}^2$

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: ${\displaystyle a^2=9\Rightarrow a=3b^2=4\Rightarrow b=2}$

Now, ${\displaystyle e=1-b^2{a}^2=1-49=53}$

Vertices are: ${\displaystyle (\pm a,0)\text{and}(0,\pm b)}$

Now, put the values of a and b to get the vertice of the conic section and we get: ${\displaystyle (\pm a,0)\text{and}(0,\pm b)=(\pm 3,0)\text{and}(0,\pm 2)}$ Now, ${\displaystyle \text{Foci}=(\pm ae,0)=(\pm 5,0)}$

Thus, the vertices and foci of the conic section are: ${\displaystyle (\pm 3,0)\text{and}(0,\pm 2)\text{and}(\pm 5,0).}$