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 $\left( {a,0} \right)$ . Show that $\dfrac{1}{{SP}} + \dfrac{1}{{SQ}}$ is independent of t.

Answer 1

With the coordinates of the given points we need to find SP and SQ using the distance formula $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$and substituting the values in $\dfrac{1}{{SP}} + \dfrac{1}{{SQ}}$we need to a value without the term t.

Complete step by step solution:
We are given two points P and Q whose coordinates are $\left( {a{t^2},2at} \right)$ and $\left( {\dfrac{a}{{{t^2}}}, - \dfrac{{2a}}{t}} \right)$
We are also given a point S whose coordinates are $\left( {a,0} \right)$
Now let's find SP and SQ
We can use the distance formula to find SQ and SP
$\Rightarrow \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$
Now we know the coordinates of S is $\left( {a,0} \right)$and P is $\left( {a{t^2},2at} \right)$
$\Rightarrow SP = \sqrt {{{\left( {a{t^2} - a} \right)}^2} + {{\left( {2at - 0} \right)}^2}} \\\ \Rightarrow SP = \sqrt {{{\left( {a{t^2} - a} \right)}^2} + {{\left( {2at} \right)}^2}} \\\ \Rightarrow SP = \sqrt {{a^2}{t^4} + {a^2} - 2{a^2}{t^2} + 4{a^2}{t^2}} \\\ \Rightarrow SP = \sqrt {{a^2}{t^4} + {a^2} + 2{a^2}{t^2}} \\\ \Rightarrow SP = \sqrt {{{\left( {a{t^2} + a} \right)}^2}} \\\ \Rightarrow SP = \left( {a{t^2} + a} \right) \\\$
Now we know the coordinates of S is $\left( {a,0} \right)$and Q is $\left( {\dfrac{a}{{{t^2}}}, - \dfrac{{2a}}{t}} \right)$
$\Rightarrow SQ = \sqrt {{{\left( {\dfrac{a}{{{t^2}}} - a} \right)}^2} + {{\left( {\dfrac{{ - 2a}}{t} - 0} \right)}^2}} \\\ \Rightarrow SQ = \sqrt {{{\left( {\dfrac{a}{{{t^2}}} - a} \right)}^2} + {{\left( {\dfrac{{ - 2a}}{t}} \right)}^2}} \\\ \Rightarrow SQ = \sqrt {\dfrac{{{a^2}}}{{{t^4}}} + {a^2} - \dfrac{{2{a^2}}}{{{t^2}}} + \dfrac{{4{a^2}}}{{{t^2}}}} \\\ \Rightarrow SQ = \sqrt {\dfrac{{{a^2}}}{{{t^4}}} + {a^2} + \dfrac{{2{a^2}}}{{{t^2}}}} \\\ \Rightarrow SQ = \sqrt {{{\left( {\dfrac{a}{{{t^2}}} + a} \right)}^2}} \\\ \Rightarrow SQ = \left( {\dfrac{a}{{{t^2}}} + a} \right) \\\$
Now we need to find the value of $\dfrac{1}{{SP}} + \dfrac{1}{{SQ}}$
Substituting the values found above we get

$$\Rightarrow \dfrac{1}{{SP}} + \dfrac{1}{{SQ}} = \dfrac{1}{{a{t^2} + a}} + \dfrac{1}{{\dfrac{a}{{{t^2}}} + a}} \\\ \Rightarrow \dfrac{1}{{SP}} + \dfrac{1}{{SQ}} = \dfrac{1}{{a{t^2} + a}} + \dfrac{1}{{\dfrac{{a + a{t^2}}}{{{t^2}}}}} \\\ \Rightarrow \dfrac{1}{{SP}} + \dfrac{1}{{SQ}} = \dfrac{1}{{a{t^2} + a}} + \dfrac{{{t^2}}}{{a{t^2} + a}} \\\ \Rightarrow \dfrac{1}{{SP}} + \dfrac{1}{{SQ}} = \dfrac{{1 + {t^2}}}{{a{t^2} + a}} \\\ \Rightarrow \dfrac{1}{{SP}} + \dfrac{1}{{SQ}} = \dfrac{{1 + {t^2}}}{{a\left( {1 + {t^2}} \right)}} \\\ \Rightarrow \dfrac{1}{{SP}} + \dfrac{1}{{SQ}} = \dfrac{1}{a} \\\$$

Now we can see that there is no t in the value of $\dfrac{1}{{SP}} + \dfrac{1}{{SQ}}$

Hence $\dfrac{1}{{SP}} + \dfrac{1}{{SQ}}$ is independent of t.

Note:
The distance formula can also be written as $\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}}$as taking the minus outside and squaring will also give the same result.
The Distance Formula is a variant of the Pythagorean Theorem