Question

Question 6 Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 49y^2 – 16x^2 = 784

Class X1 - Maths -Conic Sections Page 262

Answer 1

49y² - 16x² = 784
dividing both sides by 784 ,
49y²/784 - 16x²/784 = 784/784
y²/{784/49} - x²/{784/16} = 1
y²/16 - x²/49 = 1
y²/(4)² - x²/(7)² = 1
now, compare it with y²/a² - x²/b² = 1
so, a = 4 and b = 7

Let foci of given hyperbola are ( 0, ± c)
where , c² = a² + b²
so, c² = 4² + 7² = 16 + 49 = 65
c² = 65 => c = √65
hence, foci ( 0, ± √65)

we know, if coefficient of y² of hyperbolic equation is positive then, transverse axis is along Y-axis.
vertices = ( 0, ± a) = ( 0, ± 4)
eccentricity ( e) :
we know, when ( 0, ±c ) are Foci then, c = ae
e = c/a = √65/4
hence, eccentricity { e } = √65/4
Latusrectum = 2b²/a = 2(7)²/4 = 49/2