Question
If a hyperbola passes through the point (4, 0) and has
foci at (±5, 0), and the length of its conjugate axis is 6, then the eccentricity is:Solution
The hyperbola has foci at (±5,0), which means the center of the hyperbola is at the origin (0,0) and the major axis lies along the x-axis. The distance from the center to each focus is c. So, c=5. The length of the conjugate axis is 2b=6, which implies b=3. For a hyperbola, the relationship between a, b, and c is c2 = a2 + b2 Substituting the values of b and c:
52 = a2 + 32
a = 4 The equation of the hyperbola with center at the origin and major axis along the x-axis is The point (4,0) indeed lies on the hyperbola. This also means that the vertex of the hyperbola along the positive x-axis is at (4,0), so a=4, which is consistent with our calculation. The eccentricity e of a hyperbola is defined as e=c/a​. Substituting the values of c and a: e = 5/4
What value should come in place of (?) in the given expression.
450 ÷ 9 + 75% of 160 − 64 ÷ 4 = ?
2/5 of 3/4 of 7/9 of 14400 = ?
7/3 of 4/5 of 15/56 of ? = 83
[564 + 32 of 18 × 9 ÷ 12 + 162 ] ÷ 4 = ?
What should come in place of (?) question mark in the given expression.
 (25% of 320) + (3/8 of 400) − 30 = ?
45% of 360 - 160 + ? = √324
187 ÷ 5 ÷ 0.4 = ? – 24 × 2.4
Find the simplified value of the following expression:
[{12 + (13 × 4 ÷ 2 ÷ 2) × 5 – 8} + 13 of 8]
18 × 15 + 86 – 58 =? + 38
Â
√(24²+285-8²-172) = ?²