the equation for a hyperbola is shown

x^2/25 – y^2/49=1

what are the lengths of the transverse and conjuga

the equation for a hyperbola is shown

x^2/25 – y^2/49=1

what are the lengths of the transverse and conjugate axes?

About the author
Adalynn

1 thought on “the equation for a hyperbola is shown<br /><br /> x^2/25 – y^2/49=1<br /><br /> what are the lengths of the transverse and conjuga”

  1. Answer:

    The equation 16x^2 – 9y^2 = 144. can be written as x^2/9 – y^2/16 = – 1 This is of the form x^2/a^2 – y^2/b^2 – 1 a^2 = 9,b^2 = 16 a = 3,b = 4 Length

    Reply

Leave a Reply to Luna Cancel reply