Question

Please answer hyperbola (attached file)

Answer 1

We know that the general equation of Hyperbola is given as follows :-

`((x-h)^2)/(a^2) - ((y-k)^2)/(b^2) = 1`

Where (h,k) is the center of hyperbola.

'a' and 'b' are the length of semi-major axis and semi-minor axis respectively.

Given is the center (h,k) = (0,0); vertex (a,0) = (150,0); and some point (x,y) = (180,100)

Now we can plug the given information in the general equation of hyperbola and solve it for values of b.

`((180-0)^2)/(a^2) - ((100-0)^2)/(b^2) = 1 \\\\(32,400)/(a^2) - (10,000)/(b^2) = 1 \\\\(32,400)/((150)^2) - (10,000)/(b^2) = 1 \\\\(32,400)/(22,500) - (10,000)/(b^2) = 1 \\\\(32,400)/(22,500) - 1 = (10,000)/(b^2) \\\\(32,400-22,500)/(22,500) = (10,000)/(b^2) \\\\(9900)/(22,500) = (10,000)/(b^2) \\\\(11)/(25) = (10,000)/(b^2) \\\\\\(1)/(b^2) = (11)/(250,000)`

Hence, the equation of Hyperbola would be :-

`(x^2)/(22,500) - (11y^2)/(250,000) = 1`

Answer 2

The equation of hyperbola is

`(x^2)/(a^2)-(y^2)/(b^2)=1,` where a is the distance from the center of hyperbola to the vertex of hyperbola.

From the diagram it is seen that the center of hyperbola is at the origin O(0,0) and the vertex of hyperbola is at the point (150,0). Then the distance a=150.

Hyperbola passes through the point (180,100), so its coordinates satisfy the hyperbola equation:

`(180^2)/(150^2)-(100^2)/(b^2)=1,\\ \\(100^2)/(b^2)=(180^2)/(150^2)-1\Rightarrow (100^2)/(b^2)=(99)/(225),\\ \\b^2=(225\cdot 100^2)/(99)=(1500^2)/(99).`

Therefore, the hyperbola equation is

`(x^2)/(150^2)-(y^2)/((1500^2)/(99))=1.`