If a circle cuts a rectangular hyperbola $x y = c^{2}$ in A,B,C,D and the parameters of these four points be $t_{1}, t_{2}, t_{3}$ and $t_{4}$ respectively. Then if the eccentricity of a conic is equal to $t_{1} . t_{2} . t_{3} . t_{4}$ , the conic can't be

Circle

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Parabola

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Ellipse

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Hyperbola

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Solution

The correct options are
A Circle
C Ellipse
D Hyperbola
Let the general point on the hyperbola be $(c t, \frac{c}{t})$ .
Let the equation of the circle be $x^{2} + y^{2} + 2 g x + 2 f y + d = 0$ .
Substituting $(c t, \frac{c}{t})$ in the equation of the circle, we get, $c^{2} t^{4} + 2 g c t^{3} + d t^{2} + 2 f c t + c^{2} = 0$
Clearly, we have the product of the roots $t_{1} . t_{2} . t_{3} . t_{4} = \frac{c^{2}}{c^{2}} = 1$
Hence $e = 1 \Rightarrow$ Parabola is the required conic.
Hence correct options are $A, C, D .$

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If a circle and the rectangular hyperbola $x y = c^{2}$ meet in the four points $t_{1} . t_{2} . t_{3} & t_{4}$ then $t_{1} . t_{2} . t_{3} t_{4}$ is equal to______

Q. If a circle and the rectangular hyperbola

x y = c^{2}

meet in the four points

t_{1}, t_{2}, t_{3}

and

t_{4}

then

Q. If with standard notations

t_{1}, t_{2}, t_{3}, t_{4}

are four co–normal points on the hyperbola

x y = c^{2}

then the orthocentre of the triangle formed by joining the points

t_{1}, t_{2}, t_{3}

is given by

Q. Assertion :If points

A (c t_{1}, \frac{c}{t_{1}})

B (c t_{2}, \frac{c}{t_{2}})

C (c t_{3}, \frac{c}{t_{3}})

and

D (c t_{4}, \frac{c}{t_{4}})

lies on the circle

x^{2} + y^{2} + 2 g x + 2 f y + k = 0

then

t_{1} + t_{2} + t_{3} + t_{4} = - \frac{2 g}{c}

. Reason: Points

A, B, C, D

are parametric points of

x y = c^{2}

If a circle and rectangular hyperbola $x y = c^{2}$ meet in the four points $t_{1}, t_{2}, t_{3} & t_{4}$ . Then which of the following statements are correct?
1. The center of the mean position of the four points bisects the distance between the center of the two curve.
2. Center of the circle through the points $t_{1}, t_{2} & t_{3}$ is: $[(\frac{c}{2} (t_{1} + t_{2} + t_{3}) + \frac{1}{t_{1} . t_{2} . t_{3}}), \frac{c}{2} (\frac{1}{t_{1}} + \frac{1}{t_{2}} + \frac{1}{t_{3}} + t_{1} t_{2} t_{3})]$

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If a circle cuts a rectangular hyperbola xy=c2 in A,B,C,D and the parameters of these four points be t1,t2,t3 and t4 respectively. Then if the eccentricity of a conic is equal to t1.t2.t3.t4, the conic can't be

If a circle cuts a rectangular hyperbola $x y = c^{2}$ in A,B,C,D and the parameters of these four points be $t_{1}, t_{2}, t_{3}$ and $t_{4}$ respectively. Then if the eccentricity of a conic is equal to $t_{1} . t_{2} . t_{3} . t_{4}$ , the conic can't be