Show that the equation 9x² - 16y² - 18x + 32y - 151 = 0 represents a hyperbola. Find the co-ordinates of the centre. length of the axes, eccentricity length of latus-rectum, co-ordinates of foci and vertices, equations of the directrices of the hyperbola.
Answers
Answer:
Step-by-step explanation:
Consider the given equation
9x² - 16y² - 18x + 32y - 151 = 0
=> 9x² - 18x - 16y² + 32y - 151 = 0
=>9(x²-2x) -16(y²-2y) =151
=>9(x²-2x + 1 -1) - 16(y²-2y+1 -1) =151
=>9(x-1)²-9 -16(y-1)² + 16 = 151
=>9(x-1)² -16(y-1)² = 144
Dividing by 144 on both sides we get,
(x-1)²/16 - (y-1)²/9 = 1 which is the equation of hyperbola, whose center is given by (1, 1)
transverse axis, a = 4
conjugate axis , b = 3
eccentricity = √1 + b²/a² = √1 + 3²/4²
= 5/4.
Foci are at a distance of ae units away from the center on the transverse axis.
ae = 4*5/4 = 5
=> Foci lie on transverse axis at a distance of 5 units away from ceter
=> Foci are (1±5,1)
=(6, 1) and (-4, 1)
Vertices lie at a distance of a from center on the transverse axis.
Vertices are ( 1±4, 1)
=>(5, 1) and (-3, 1)
Directrices are the lines which are perpendicular to transverse axis and at a distance of a/e from center
=> Any point at a distance of a/e from center is given by
(1 ± 4/5/4, 1)
= (21/5, 1 ) and (-11/5, 1).
Hence equations of directrices, which are perpendicular to transverse axis (y - 1= o ) will be of the form x = const but this passes through (21/5, 1 ) and (-11/5, 1).Hence equations are x = 21/5 and x = -11/5.