SATHEE: Hyperbola 2 Question 13

Hyperbola 2 Question 13

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)

13.

Equation of the circle with $A B$ as its diameter is

(a) $x^{2} + y^{2} - 12 x + 24 = 0$

(b) $x^{2} + y^{2} + 12 x + 24 = 0$

(d) $x^{2} + y^{2} - 24 x - 12 = 0$

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Answer:

Correct Answer: 13. (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, $(x - 6)^{2} + (y + 2 \sqrt{3}) (y - 2 \sqrt{3}) = 0$

$\begin{array}{ll} \Rightarrow & x^{2} - 12 x + y^{2} + 24 = 0 \\ \Rightarrow & x^{2} + y^{2} - 12 x + 24 = 0 \end{array}$

Hyperbola 2 Question 13

13.

Answer:

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Hyperbola 2 Question 13

13.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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