Assertion(A): lf the lines $3 x + y + p = 0$ and $2 x + 5 y - 3 = 0$ are conjugate with respect to $3 x^{2} - 2 y^{2} = 6$ then $p = 1$
Reason(R): lf the lines $l_{1} x + m_{1} y + n_{1} = 0$ and $l_{2} x + m_{2} y + n_{2} = 0$ are conjugate with respect to the hyperbola $S = 0$ is $a^{2} l_{1} l_{2} + b^{2} m_{1} m_{2} = n_{1} n_{2}$

Both A and R are true and R is the correct
explanation of A.

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Both A and R are true but R is not correct
explanation of A.

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A is true but R is false

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A is false but R is true

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Solution

The correct option is D A is true but R is false
lf the lines $l_{1} x + m_{1} y + n_{1} = 0$ and $l_{2} x + m_{2} y + n_{2} = 0$ are conjugate with re- spect to the hyperbola $S = 0$ is $a^{2} l_{1} l_{2} - b^{2} m_{1} m_{2} = n_{1} n_{2}$
then $2 (3) (2) - 3 (1) (5) = - 3 p$
therefore, $p = 1$

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Similar questions

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l_{1} x + m_{1} y + n_{1} = 0

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Assertion(A): lf the lines 3x+y+p=0 and 2x+5y−3=0 are conjugate with respect to 3x2−2y2=6 then p=1Reason(R): lf the lines l1x+m1y+n1=0 and l2x+m2y+n2=0 are conjugate with respect to the hyperbola S=0 is a2l1l2+b2m1m2=n1n2

The correct option is D A is true but R is false lf the lines l1x+m1y+n1=0 and l2x+m2y+n2=0 are conjugate with re- spect to the hyperbola S=0 is a2l1l2−b2m1m2=n1n2then 2(3)(2)−3(1)(5)=−3ptherefore, p=1

The correct option is D A is true but R is false
lf the lines $l_{1} x + m_{1} y + n_{1} = 0$ and $l_{2} x + m_{2} y + n_{2} = 0$ are conjugate with re- spect to the hyperbola $S = 0$ is $a^{2} l_{1} l_{2} - b^{2} m_{1} m_{2} = n_{1} n_{2}$
then $2 (3) (2) - 3 (1) (5) = - 3 p$
therefore, $p = 1$