Question

Coordinate geometry/distance formula...solve it step by step..

Answer 1

When there is root, take as much things common as you can, so that you can write them separately and take root of perfect squares if it is there.

I am just solving the under root part which is equal to AB. I am not writing AB each time.

I am taking t1 as x and t2 as y. it will be easier to write. I will convert them to t1 and t2 at the end.

$\sqrt{{[a {y}^{2} - a {x}^{2}]}^{2} + {[2ay - 2ax]}^{2} }$

$= \sqrt{{[a ({y}^{2} - {x}^{2})]}^{2} + {[2a(y - x)]}^{2} }$

$= \sqrt{{{a}^{2} ({y}^{2} - {x}^{2})}^{2} + {4 {a}^{2} (y - x)}^{2} }$

$= \sqrt{{{a}^{2} [(y + x)(y - x)]}^{2} + {4 {a}^{2} (y - x)}^{2} }$

$= \sqrt{{{a}^{2} (y + x)^{2} (y - x)}^{2} + {4 {a}^{2} (y - x)}^{2} }$

$= \sqrt{{({a}^{2} ) \times (y - x)}^{2} \times ( (y + x)^{2} \: + {4}) }$

$= \sqrt{ {a}^{2} } \times \sqrt{ {(y - x)}^{2} } \times \sqrt{ (y + x)^{2} \: + {4} }$

$= a(y - x) \times \sqrt{ ({y + x})^{2} + 4}$

$= a(t_2 - t_1) \times \sqrt{ ({t_2 + t_1})^{2} + 4}$

$so \: \: AB = a(t_2 - t_1) \times \sqrt{ ({t_2 + t_1})^{2} + 4}$