Question

How do you identify the vertices, foci and direction of \[dfrac{{{{(x - 3)}^2}}}{4} - dfrac{{{{(y + 1)}^2}}}{9} = 1\]?

Answer 1

Hint: First, we will try to find the center of the parabola. Then, we will find the vertex by converting the negative term of the equation to zero. Then, we will simplify the term and get the answer. Then, we will find the foci by applying the formula of foci.

Complete step-by-step solution:
According to the question, we will first take out the center of the parabola. The center of parabola here is \[(3, - 1)\].
Now, we will find the vertices. So, first, we will try to convert the negative term from the equation to zero. Here, the negative term is \[dfrac{{{{(y + 1)}^2}}}{9}\]. We will convert it to 0 by taking \[y = - 1\], and we get:
\[ \Rightarrow dfrac{{{{(x - 3)}^2}}}{4} - dfrac{{{{( - 1 + 1)}^2}}}{9} = 1\]
\[ \Rightarrow dfrac{{{{(x - 3)}^2}}}{4} - 0 = 1\]
\[ \Rightarrow dfrac{{{{(x - 3)}^2}}}{4} = 1\]
Now, we will try to shift the denominator to the other side of the equation, and we get:
\[ \Rightarrow {(x - 3)^2} = 4\]
This equation can also be written as:
\[ \Rightarrow {(x - 3)^2} = {(2)^2}\]
Now, we can take out the square roots from both the sides. This gives us two equations, where in one equation, we will have a positive 2, and the other equation will have a negative 2:
\[x - 3 = 2\]
\[x - 3 = - 2\]
After simplifying the first equation \[x - 3 = 2\], we get:
\[ \Rightarrow x = 5\]
After simplifying the second equation, we get:
\[ \Rightarrow x = 1\]
Therefore, the vertices we got are:
\[(1, - 1)\,\,and\,\,(5, - 1)\]
Now, we will find the foci. We will take focus here as ‘c’. We know that -1 is the y coordinate for both foci. The coordinate x will be either plus or minus the distance ‘c’ from the x coordinate of the centre 3.
The formula for ‘c’ is:
\[{c^2} = {a^2} + {b^2}\]
Here, \[{a^2},{b^2}\]are the values which were in the denominators of the equation.
When we put the values of the denominator in the formula, we get:
\[ \Rightarrow {c^2} = 4 + 9\]
\[ \Rightarrow {c^2} = 13\]
\[ \Rightarrow c = \sqrt {13} \]

Therefore, the foci are \[(3 - \sqrt {13} , - 1)\,\,and\,\,(3 + \sqrt {13} , - 1)\]

Note: A hyperbola is a curve which is open containing two branches, the intersection of a plane with both halves of a double one. The foci of a hyperbola are two points which are fixed. The points are inside each curve inside the hyperbola.