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

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.