Question

derive the equation for hyperbola

Answer 1

Answer:

https://courses.lumenlearning.com/ivytech-collegealgebra/chapter/deriving-the-equation-of-a-hyperbola-centered-at-the-origin/

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Step-by-step explanation:

Answer 2

Step-by-step explanation:

take a point P(x, y) on the hyperbola such that, PF1 – PF2 = 2a

By the distance formula, we have,

√ {(x + c)2 + y2} – √ {(x – c)2 + y2} = 2a

Or, √ {(x + c)2 + y2} = 2a + √ {(x – c)2 + y2}

Further, let’s square both the sides. Hence, we have

(x + c)2 + y2 = 4a2 + 4a√ {(x – c)2 + y2} + (x – c)2 + y2

On simplifying the equation, we get

√ {(x – c)2 + y2} = x(c/a) – a

We square both sides again and simplify it further to get,

x2/a2 – y2/(c2 – a2) = 1

We know that c2 – a2 = b2. Therefore, we have

x2/a2 – y2/b2 = 1

Therefore, we can say that any point on the hyperbola satisfies the equation:

x2/a2 – y2/b2 = 1 … (1)

Let’s look at the converse situation now. If P(x, y) satisfies equation (1) with 0 < a < c, then

y2 = b2{(x2 – a2)/a2}

Therefore, PF1 = √ {(x + c)2 + y2}

= √ {(x + c)2 + b2[(x2 – a2)/a2])}

On simplifying the equation, we get PF1 = a + x(c/a)

Using similar calculations for PF2, we get PF2 = a – x(c/a)

In hyperbola c > a; and since P is to the right of the line x = a, x > a, and (c/a)x > a. Therefore, a – (c/a)x becomes negative.

Thus, PF2 = (c/a)x – a

Therefore, PF1 – PF2 = {a + x(c/a)} – {x(c/a) – a} = a + x(c/a) – x(c/a) + a = 2a.

Also, note that if P is to the left of the line x = – a, then,

PF1 = – {a + (c/a)x} and PF2 = a – (c/a)x

In this case, PF1 – PF2 = 2a.

Hence, it is evident that any point that satisfies the equation x2/a2 – y2/b2 = 1, lies on the hyperbola.