Question

the equation of the hyperbola that has a center at (7,3), a focus at (2,3), and a vertex at (10,3), is
\\( \frac { ( x - c ) ^ { 2 } } { a ^ { 2 } } - \frac { ( y - d ) ^ { 2 } } { b ^ { 2 } } = 1 \\)
where
\\( a = \\)
\\( b = \\)
\\( c = \\)
\\( d = \\)
question help: video

Explanation:

Step1: Find \(C\) and \(D\)

The center of the hyperbola is \((h,k)=(C,D)=(7,3)\) (from the general form of hyperbola center \((h,k)\)).

Step2: Find \(A\)

The distance between the center \((7,3)\) and a vertex \((10,3)\) is given by \(A=\vert x - h\vert\) (since the \(y -\)coordinates of center and vertex are the same, it's a horizontal - transverse axis hyperbola). Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), with \((x_1,y_1)=(7,3)\) and \((x_2,y_2)=(10,3)\), we have \(A=\vert10 - 7\vert=3\).

Step3: Find \(c\)

The distance between the center \((7,3)\) and a focus \((2,3)\) is \(c=\vert x - h\vert\). Using the distance formula with \((x_1,y_1)=(7,3)\) and \((x_2,y_2)=(2,3)\), we get \(c=\vert2 - 7\vert = 5\).

Step4: Find \(B\)

For a hyperbola, the relationship \(c^{2}=A^{2}+B^{2}\). We know \(A = 3\) and \(c = 5\). Substitute into the formula: \(B^{2}=c^{2}-A^{2}\). Then \(B^{2}=25 - 9=16\), so \(B = 4\).

Answer:

\(A = 3\), \(B = 4\), \(C = 7\), \(D = 3\)