Question

if A=(at^2,2at);B(a/t^2,-2a/t) and s(a,0) then 1/SA+1/SB​

Answer 1

Step-by-step explanation:

SQ = \sqrt{(a- \frac{a}{t^2})^2 + (0+ \frac{2a}{t})^2}

= \sqrt{(\frac{t^2a-a}{t^2})^2 + (\frac{2a}{t})^2}

= \sqrt{ \frac{t^4a^2+a^2-2a^2t^2}{t^4} + \frac{4a^2}{t^2} }

= \frac{t^4a^2 + a^2 - 2a^2t^2 + 4a^2t^2}{t^4}

= \sqrt{ \frac{t^4a^2 + 2a^2t^2 + a^2 }{t^4} }

= \sqrt{ \frac{(t^2a+a)^2}{t^4} }

= \frac{t^2a+a}{t^2}

\frac{1}{SQ} = \frac{t^2 }{t^2a+a} ..........(1)

Now,

SP = \sqrt{(a-at^2)^2 + (0-2at)^2 }

= \sqrt{a^2 - 2a^2t^2 + a^2t^4 + 4a^2t^2}

= \sqrt{a^2 + 2a^2t^2 + a^2t^4}

= \sqrt{(at^2+a)^2}

= at² + a

\frac{1}{SP} = \frac{1}{at^2 + a}

\frac{1}{SQ} + \frac{1}{SP} = \frac{t^2 }{t^2a+a}+\frac{1}{at^2 + a}

= \frac{t^2+1}{a(t^2+1)}

= \frac{1}{SQ} + \frac{1}{SP} = \frac{1 }{a}

Answer 2

SQ = \sqrt{(a- \frac{a}{t^2})^2 + (0+ \frac{2a}{t})^2}

= \sqrt{(\frac{t^2a-a}{t^2})^2 + (\frac{2a}{t})^2}

= \sqrt{ \frac{t^4a^2+a^2-2a^2t^2}{t^4} + \frac{4a^2}{t^2} }

= \frac{t^4a^2 + a^2 - 2a^2t^2 + 4a^2t^2}{t^4}

= \sqrt{ \frac{t^4a^2 + 2a^2t^2 + a^2 }{t^4} }

= \sqrt{ \frac{(t^2a+a)^2}{t^4} }

= \frac{t^2a+a}{t^2}

\frac{1}{SQ} = \frac{t^2 }{t^2a+a} ..........(1)

Now,

SP = \sqrt{(a-at^2)^2 + (0-2at)^2 }

= \sqrt{a^2 - 2a^2t^2 + a^2t^4 + 4a^2t^2}

= \sqrt{a^2 + 2a^2t^2 + a^2t^4}

= \sqrt{(at^2+a)^2}

= at² + a

\frac{1}{SP} = \frac{1}{at^2 + a}

\frac{1}{SQ} + \frac{1}{SP} = \frac{t^2 }{t^2a+a}+\frac{1}{at^2 + a}

= \frac{t^2+1}{a(t^2+1)}

= \frac{1}{SQ} + \frac{1}{SP} = \frac{1 }{a}