Question

If P and Q are two points whose co-ordinates are (at square,2at) &(a/t square,2a/t) respectively &S is in the point (a,0). Show that 1/Sp + 1/SQ is independent of't'? plzzzz answer this immediately it's urgent tomorrow is my exam ​

Answer 1

Answer:

Refers to attachmnets...

Answer 2

Solution

P(at²,2at)

Q(a/t²,2a/t)

and

S(a,0)

now...

$sp = \sqrt{(a - at {}^{2}) {}^{2} + (0 - 2at) {}^{2} } \\ = > sp = \sqrt{a {}^{2} (1 - t {}^{2} ) {}^{2} + a {}^{2}(4t {}^{2} ) } \\ = > sp = a \sqrt{1 + t {}^{4} - 2t {}^{2} + 4t {}^{2} } \\ = > sp =a \sqrt{1 + 2t {}^{2} + t {}^{4} } \\ = > sp = a \sqrt{(1 + t {}^{2} ) {}^{2} } \\ = > sp = a(1 + t {}^{2} )$

and..

$sq = \sqrt{(a - \frac{a}{t {}^{2} }) {}^{2} + (0 - \frac{2a}{t}) {}^{2} } \\ = > sq = a \sqrt{ \frac{(t {}^{2} - 1) {}^{2} }{t {}^{4} } + \frac{4}{t {}^{2} } } \\ = > sq = a \sqrt{ \frac{t {}^{4} + 1 - 2t {}^{2} + 4t {}^{2} }{t ^{4} } } \\ = > sq = \frac{a}{t {}^{2} } \sqrt{(1 + t {}^{2} ) {}^{2} } \\ = > sq = \frac{a(1 + t {}^{2}) }{t {}^{2} }$

now....

$= \frac{1}{sp} + \frac{1}{sq} \\ = \frac{1}{a(1 + t {}^{2} )} + \frac{t {}^{2} }{a(1 + t {}^{2}) } \\ = \frac{(1 + t {}^{2}) }{a(1 + t {}^{2}) } \\ = \frac{1}{a} \\ i.e. \: independent \: of \: t \: (proved)$

hope this helps you ✌️