Question

b^2=a^2(1-e^2) please derive​

Answer 1

Derivation of the Equation

Now, we take a point P(x, y) on the ellipse 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

Simplifying the equation, we get √ {(x – c)2 + y2} = a – x(c/a)

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

x2/a2 + y2/(a2 – c2) = 1

We know that c2 = a2 – b2. Therefore, we have x2/a2 + y2/b2 = 1

Therefore, we can say that any point on the ellipse 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 < c < a, then y2 = b2(1 – x2/a2)

Therefore, PF1 = √ {(x + c)2 + y2}

= √ {(x + c)2 + b2(1-x2/a2)}

Simplifying the equation and replacing b2 with a2 – c2, we get PF1 = a + x(c/a)

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

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

Therefore, any point that satisfies equation (1), i.e. x2/a2 + y2/b2 = 1, lies on the ellipse. Also, the equation of an ellipse with centre of the origin and major axis along the x-axis is:

x2/a2 + y2/b2 = 1.

Note: Solving the equation (1), we get

x2/a2 = 1 – y2/b2 ≤ 1

Therefore, x2 ≤ a2. So, – a ≤ x ≤ a. Hence, we can say that the ellipse lies between the lines x = – a and x = a and touches these lines. Similarly, it can lie between the lines y = – b and y = b and touch those lines. Its equation {Fig. 5 (b)} is:

x2/b2 + y2/a2 = 1.

Hence the Standard Equations of Ellipses are:

x2/a2 + y2/b2 = 1.

x2/b2 + y2/a2 = 1.