Question

If a, b, c form a G.P. with a common ratio r. P and Q are two points whose coordinates satisfy the relation \[ax + by + c = 0\] and \[{x^2} - {y^2} - 4 = 0\] then ….
(This question has multiple correct options)
A. Sum of the ordinates of P and Q is \[\dfrac{{2{r^3}}}{{1 - {r^2}}}\].
B. Sum of the ordinates of P and Q is \[\dfrac{{2{r^3}}}{{{r^2} - 1}}\].
C. Product of the ordinates of P and Q is \[\dfrac{{{r^4} - 4}}{{{r^2} - 1}}\].
D. Product of the ordinates of P and Q is \[\dfrac{{4 - {r^4}}}{{{r^2} - 1}}\].

Answer 1

Hint: Use the concept of G.P. in G.P. consecutive terms are having a common ratio in them. Also use the relation between roots of a quadratic equation.

Complete step-by-step answer:
\[ax + by + c = 0\] …..equation1
\[{x^2} - {y^2} - 4 = 0\] …….equation2
First it is given that a, b, c are in G .P. with common ratio r.
Then,
a is the first term.
b= \[ar\]
c= \[a{r^2}\]
putting these values in given equation1,
\[ax + ary + a{r^2} = 0\]
Taking a common,
\[x + ry + {r^2} = 0\]
\[x = - \left( {ry + {r^2}} \right)\]
Now putting this value in equation2
\[
  {\left( { - \left( {ry + {r^2}} \right)} \right)^2} - {y^2} - 4 = 0 \\
  {(ry)^2} + 2{r^3}y + {({r^2})^2} - {y^2} - 4 = 0 \\
  {r^2}{y^2} + 2{r^3}y + {r^4} - {y^2} - 4 = 0 \\
  \left( {{r^2} - 1} \right){y^2} + 2{r^3}y + {r^4} - 4 = 0 \\
\]
Now ordinates of P and Q are roots of,
\[\left( {{r^2} - 1} \right){y^2} + 2{r^3}y + {r^4} - 4 = 0\]
Thus relation can be defined as
Sum of roots =\[\dfrac{{ - 2{r^3}}}{{{r^2} - 1}}\]
Product of roots = \[\dfrac{{{r^4} - 4}}{{{r^2} - 1}}\]
So in option A, taking minus common from terms of denominator we will get \[\dfrac{{ - 2{r^3}}}{{{r^2} - 1}}\]
So option A and C are the correct answers.


Note: Students can get confused in option A and B due to negative signs. In option A we need negative signs to be taken common from denominator .Then it will match your answer. So choose options wisely.