Answer:
Length of the transverse axis of the given conic section = 8 units.
Step-by-step explanation:
The equation of hyperbola that has a horizontal orientation in the standard form is;
\(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) .....[1]
The center is (h, k) , the vertices are \((h \pm a , k)\) and \((h , k\pm b)\)
Since, the vertices are on the hyperbola, but the co-vertices are not on the hyperbola.
- The conjugate axis is the line segment with endpoints at the co-vertices and has length 2b
- The transverse axis is the line segment with endpoints at the vertices and has length 2a.
Since, the equation of the conic section is given as:
\(\frac{(x+2)^2}{16} - \frac{(y-3)^2}{9} = 1\)
On comparing with equation [1]
⇒ center = ( -2 , 3 ) and
\(a^2 = 16\)
\(a = \sqrt{16} = 4\) units
and
\(b^2 = 9\)
\(b = \sqrt{9} = 3\) units
Length of the transverse axis is the distance between the two vertices, 2a = 2(4) = 8 units.
