Question

Find the standard equation of the conic with center at (-7,-3), one of its vertices at (-7,4), and one focus at (-7,3).

Answer 1

A conic with center, vertex and focus is an Elispse

Write out the givencoordinates

`\begin{gathered} \text{center}=(-7,-3) \\ \text{vertex}=(-7,4) \\ \text{Focus =(-7,3)} \end{gathered}`

Since the x-coordinate of the given vertices and foci are the same, the equation becomes

`(\mleft(x-h\mright)^2)/(b^2)+((y-k)^2)/(a^2)=1`

Where

`\begin{gathered} \text{The center is (h,k) coordinates of the center } \\ \text{hence } \\ h=-7,k=-3 \end{gathered}`

Then

`\begin{gathered} \text{The vertices are } \\ (-7,4)\text{ and the other vertex is } \\ (-7,-10) \\ \end{gathered}`

Hence

`\begin{gathered} \text{Distance betw}en\text{ the two given} \\ 4-(-10)=4+10=14 \\ \text{Then} \\ 2a=14 \\ a=(14)/(2)=7 \end{gathered}`

Then

From the given coordinates of the focus (h,k±c) and center (h,k)

`\begin{gathered} \text{focus =(-7,3)} \\ \text{Center}=(-7,-3) \\ \text{then } \\ K=-3, \\ K+C=3 \\ \text{Then} \\ -3+C=3 \\ C=3+3=6 \end{gathered}`

hence

`\begin{gathered} a=7,a^2=7^2=49 \\ c=6^2,c^2=6^2=36 \end{gathered}`

Using

`\begin{gathered} c^2=a^2-b^2 \\ \text{Substitute the value } \\ 36=49-b^2 \\ 36-49=-b^2 \\ \text{Multiply throught by -1} \\ b^2=49-36 \\ b^2=13 \end{gathered}`

Recall the standard equation as

`\begin{gathered} ((x-h)^2)/(b^2)+((y-k)^2)/(a^2)=1 \\ h=-7,k=-3,b^2=13,a^2=49 \\ \text{Then } \\ ((x-(-7))^2)/(13)+((y-(-3))^2)/(49)=1 \\ \end{gathered}`

Therefore

The standard equation becomes

`((x+7)^2)/(13)+((y+3)^2)/(49)=1`

Therefore the standard equation is of the conic is (x+7)²/13 + (y+3)²/49=1