Hyperbola Ques 23
Passage Based Problems
The circle $x^{2}+y^{2}-8 x=0$ and hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$ intersect at the points $A$ and $B$.
(2010)
Equation of the circle with $A B$ as its diameter is
(a) $x^{2}+y^{2}-12 x+24=0$
(b) $x^{2}+y^{2}+12 x+24=0$
(c) $x^{2}+y^{2}+24 x-12=0$
(d) $x^{2}+y^{2}-24 x-12=0$
Show Answer
Answer:
Correct Answer: 23.(a)
Solution:
- The equation of the hyperbola is $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$ and that of circle is $x^{2}+y^{2}-8 x=0$
For their points of intersection, $\frac{x^{2}}{9}+\frac{x^{2}-8 x}{4}=1$
$ \begin{array}{lr} \Rightarrow & 4 x^{2}+9 x^{2}-72 x=36 \\ \Rightarrow & 13 x^{2}-72 x-36=0 \\ \Rightarrow & 13 x^{2}-78 x+6 x-36=0 \\ \Rightarrow & 13 x(x-6)+6(x-6)=0 \\ \Rightarrow & x=6, x=-\frac{13}{6} \end{array} $
$x=-\frac{13}{6}$ not acceptable.
Now, for $x=6, y= \pm 2 \sqrt{3}$
Required equation is, $(x-6)^{2}+(y+2 \sqrt{3})(y-2 \sqrt{3})=0$
$ \begin{array}{ll} \Rightarrow & x^{2}-12 x+y^{2}+24=0 \\ \Rightarrow & x^{2}+y^{2}-12 x+24=0 \end{array} $
BETA
