Question

Prove that the distance between (at*2,2at) and(a/t*2,-2a/t) is a(t+1/t)*2

Answer 1

Given points are 
A (at²,2at) , B (a/t²,-2a/t)
Now we calculate the distance between these points by using the distance formula.
S = √[(at²-a/t²)² + (2at+2a/t)²]
   = √[a²(t²-1/t²)² + 4a²(t+1/t)²]
   = √[a²(t⁴+1/t⁴-2) + 4a²(t²+1/t²+2)]
   = a √(t⁴+1/t⁴-2+4t²+4/t²+8)
   = a √ (t⁴+4t²+6+4/t²+1/t⁴)
   = a √ (t + 1/t)⁴
   = a √ [(t + 1/t)²]²
   = a (t + 1/t)²
This is the required answer.
Thanks.

Answer 2

Question: Prove that the distance between (at², 2at) and (a/t², -2a/t) is

a(t + 1/t)².

Answer:

The distance between (at², 2at) and (a/t², -2a/t) is a(t + 1/t)² is proved.

Step-by-step explanation:

Given:-

The points are (at², 2at) and (a/t², -2a/t).

To prove:-

The distance between (at², 2at) and (a/t², -2a/t) is a(t + 1/t)².

Step 1 of 1

Consider the given two points as follows:

A = (at², 2at)

B = (a/t², -2a/t)

or,

$(x_1,y_1)$ = (at², 2at)

$(x_2,y_2)$ = (a/t², -2a/t)

Then,

The distance between the points A and B is,

AB = $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

AB = $\sqrt{\left(at^2-\frac{a}{t^2}\right)^2 + \left(2at-\frac{-2a}{t}\right)^2}$

     = $\sqrt{\left(at^2-\frac{a}{t^2}\right)^2 + \left(2at+\frac{2a}{t}\right)^2}$

Using the identity, (a ± b)² = a² + b² ± 2ab

AB = $\sqrt{\left(a^2t^4+\frac{a^2}{t^4}-2a^2\right) + \left(4a^2t^2+\frac{4a^2}{t^2}+8a^2\right)}$

     = $\sqrt{a^2\left(t^4+\frac{1}{t^4}-2 + 4t^2+\frac{4}{t^2}+8\right)}$

     = $a\sqrt{\left(t^4+\frac{1}{t^4} + 4t^2+\frac{4}{t^2}+6\right)}$

Further, simplify as follows:

AB = $a\sqrt{\left(\frac{t^8+1+4t^6+4t^2+6t^4}{t^4} \right)}$

     = $a\sqrt{\left(\frac{t+1}{t} \right)^4}$

AB = $a\left(\frac{t+1}{t} \right)^2$

AB = $a\left(1+\frac{1}{t} \right)^2$

Therefore, the distance between (at², 2at) and (a/t², -2a/t) is a(t + 1/t)².

#SPJ3