Question

If P and Q are two points whose coordinates are $ \left( {a{t^2},2at} \right) $ and $ \left( {\dfrac{a}{{{t^2}}},\dfrac{{2a}}{t}} \right) $ respectively and S is the point $ (a,0) $ . Show that $ \dfrac{1}{{SP}} + \dfrac{1}{{SQ}} $ is independent of it.

Answer 1

Hint: From the given points find the length or the distance among the two points using the distance formula. Then substitute the values in the given condition and simplify the equation to check whether the resultant values are expressed in terms of “t” or not.

Complete step-by-step answer:
Here we are given three points-
 $ P = \left( {a{t^2},2at} \right) $
 $ Q = \left( {\dfrac{a}{{{t^2}}},\dfrac{{2a}}{t}} \right) $
 $ S = (a,0) $
First find the length of SP and SQ using the distance formula.
 $ SP = \sqrt {{{(a - a{t^2})}^2} + {{(0 - 2at)}^2}} $
Expand the square of the terms and simplify the above equation –
 $ SP = \sqrt {(a{}^2 - 2{a^2}{t^2} + {a^2}{t^4} + 4{a^2}{t^2})} $
Simplify the addition and subtraction in the pair of the like terms-
\[
  SP = \sqrt {(a{}^2 + {a^2}{t^4} + \underline {4{a^2}{t^2} - 2{a^2}{t^2}} )} \\
  SP = \sqrt {a{}^2 + {a^2}{t^4} + 2{a^2}{t^2}} \\
 \]
The above equation can be simplified by using - $ {(a + b)^2} = {a^2} + 2ab + {b^2} $
 $ \Rightarrow SP = \sqrt {{{(a + a{t^2})}^2}} $
Square and square-root cancel each other on the right hand side of the equation –
 $ \Rightarrow SP = (a + a{t^2})\;{\text{ }}.....{\text{ (i)}} $
Similarly,
 $ SQ = \sqrt {{{(a - \dfrac{a}{{{t^2}}})}^2} + {{(0 - \dfrac{{2a}}{t})}^2}} $
Expand the square of the terms and simplify the above equation –
 $ SQ = \sqrt {(a{}^2 - \dfrac{{2{a^2}}}{{{t^2}}} + \dfrac{{{a^2}}}{{{t^4}}} + \dfrac{{4{a^2}}}{{{t^2}}})} $
Simplify the addition and subtraction in the pair of the like terms-
 $ SQ = \sqrt {\left( {a{}^2 + \dfrac{{{a^2}}}{{{t^4}}} + \underline {\dfrac{{4{a^2}}}{{{t^2}}} - \dfrac{{2{a^2}}}{{{t^2}}}} } \right)} $
 $ SQ = \sqrt {\left( {a{}^2 + \dfrac{{{a^2}}}{{{t^4}}} + \dfrac{{2{a^2}}}{{{t^2}}}} \right)} $
The above equation can be simplified by using - $ {(a + b)^2} = {a^2} + 2ab + {b^2} $
 $ \Rightarrow SQ = \sqrt {{{\left( {a + \dfrac{a}{{{t^2}}}} \right)}^2}} $
Square and square-root cancel each other on the right hand side of the equation –
 $ \Rightarrow SQ = \left( {a + \dfrac{a}{{{t^2}\;}}} \right){\text{ }}.....{\text{ (ii)}} $
By using the equations (i) and (ii), place the values in the given condition –
 $ \dfrac{1}{{SP}} + \dfrac{1}{{SQ}} = \dfrac{1}{{a + a{t^2}}} + \dfrac{1}{{a + \dfrac{a}{{{t^2}}}}} $
Simplify the above equation by making the denominators the same.
 $ \dfrac{1}{{SP}} + \dfrac{1}{{SQ}} = \dfrac{1}{{a + a{t^2}}} + \dfrac{{{t^2}}}{{a + a{t^2}}} $ [Denominator’s denominator goes to the numerator]
Directly add the numerator, when the denominator is same for both the fraction –
 $ \dfrac{1}{{SP}} + \dfrac{1}{{SQ}} = \dfrac{{1 + {t^2}}}{{a + a{t^2}}} $
Take common factors and simplify –
 $ \dfrac{1}{{SP}} + \dfrac{1}{{SQ}} = \dfrac{{(1 + {t^2})}}{{a(1 + {t^2})}} $
Common multiple from the numerator and the denominator cancel each other-
 $ \dfrac{1}{{SP}} + \dfrac{1}{{SQ}} = \dfrac{1}{a} $
The above expression shows that the resultant value is independent of “t”.
Hence, proved.

Note: One should be very clear in the concepts of the LCM (Least common Multiple) and the simplification of the fractions and its properties. Always remember for addition or subtraction between any fractions, they should have the same denominators.