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Question: Answered & Verified by Expert
If the circle $x^2+y^2=a^2$ intersects the hyperbola $x y=b^2$ at four points $\left(x_1, y_1\right)$, $\left(x_2, y_2\right),\left(x_3, y_3\right),\left(x_4, y_4\right)$, then $y_1 y_2 y_3 y_4=$
MathematicsHyperbolaTS EAMCETTS EAMCET 2020 (10 Sep Shift 2)
Options:
  • A $a^4$
  • B 0
  • C $b^4$
  • D $b^2$
Solution:
2475 Upvotes Verified Answer
The correct answer is: $b^4$
We have,
$x^2+y^2=a^2$ $\ldots(\mathrm{i})$
and $x y=b^2$ $\ldots(\mathrm{ii})$


Let four points $A\left(x_1, y_1\right), B\left(x_2, y_2\right), C\left(x_3, y_3\right)$ and $D\left(x_4, y_4\right)$
$\therefore \quad O A^2=x_1^2+y_1^2$
$O B^2=x_2^2+y_2^2$
$O C^2=x_3^2+y_3^2$
and $\quad O D^2=x_4^2+y_4^2$
Such that $O A^2+O B^2+O C^2+O D^2$
$=\Sigma x_1^2+\Sigma y_1^2=4 a^2$
From Eq. (ii) $x=\frac{b^2}{y}$
From Eq. (i) $\left(\frac{b^2}{y}\right)^2+y^2=a^2$
$\Rightarrow \quad b^4+y^4=a^2 y^2$
$\Rightarrow \quad y^4-a^2 y^2+b^4=0$
This is an equation of 4 th degree in $y$ and its four roots are $y_1, y_2, y_3, y_4$, then $y_1 y_2 y_3 y_4=b^4$

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