Draw the In Exercises 114 you are asked to find the equation

Draw the In Exercises 1-14 you are asked to find the equation of a hyperbola, based on the given information.

Solution

Find Hyperbola Equation,

14th Question Where Vertices V(-7,1) and V’ (1,1) and Eccentricity = 3

Answer)   From vertices, we can see that Hyperbola lies on Side by side on Y-Axis

So the y part of the equation will be subtracted and the a² will go with the x part of the equation

Vertices over X-axis are at -7 and 1, so at a distance of 8 points.

Hence, Centre is at (-3,1)

A² = 4² = 16

Eccentricity = c/a = 3

using the a² + b² = c² formula, we get

b square = 12² – 4² = 144- 16 = 128

We get the Hyperbola equation as:

(x-h)²/ a²     - (y-k)²/b²    = 1

(x+3)²/ 16 - (y-1)²/128 = 1

12^th Answer)

Asymptote y + 2 = ±3 (x-4) and one vertex V at (4, 5)

Asymptote equation are

Y+2 = 3(x-4) and Y+2 = -3(x-4)

Both these lines intersect at Centre and then values of x and y are same in both the equation.

Hence, equating both the equations we get:

Applying this value, we get value of Y as -2

Hence, Centre of the equation is (4,-2)

One of the Vertex is at 4,5, so Hyperbola is above and below to each other parallel to Y axis. Then the a² will go with the y part of the hyperbola equation, and the x part will be subtracted.

Distance between Centre and Vertex along Y axis is 5 –(-2) = 7

So, a = 7

Y = mx+c where Slope is M for Asymptote line.

Hence, Slope M = 3 = b/a

Hyperbola’s equation is given as

(y-k)² / a² – (x-h)² / b² = 1

(Y+2)² / 49 – (x-4)² / 441 = 1