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Question: If \[{\text{xy}} - 4{\text{x}} + 3{\text{y}} - {{\lambda }} = 0\] represents the asymptotes of \[{\t...

If xy4x+3yλ=0{\text{xy}} - 4{\text{x}} + 3{\text{y}} - {{\lambda }} = 0 represents the asymptotes of xy4x+3y=0{\text{xy}} - 4{\text{x}} + 3{\text{y}} = 0, then find the value ofλ{{\lambda }}.
A{\text{A}}. 33
B{\text{B}}. 6 - 6
C{\text{C}}. 88
D{\text{D}}. 1212

Explanation

Solution

From the given question, we have to find the value of λ{{\lambda }} and choose the correct answer. First, we have to find the joint equation of asymptotes by the given equation then we have to compare the joint equation to the given asymptotic equation, we get the required result.

Complete step-by-step solution:
We have to find the value of λ{{\lambda }} by the given equation xy4x+3y=0{\text{xy}} - 4{\text{x}} + 3{\text{y}} = 0.
First, we are going to find the joint equation of asymptotes.
Given that, xy4x+3y=0{\text{xy}} - 4{\text{x}} + 3{\text{y}} = 0.
Now, add and subtract the term 1212 on the left hand side (LHS) of the above term. Then, we get
xy4x+3y12+12=0\Rightarrow {\text{xy}} - 4{\text{x}} + 3{\text{y}} - 12 + 12 = 0
Let us separate the common term on the left hand side (LHS)
x(y4)+3(y4)+12=0\Rightarrow {\text{x}}\left( {{\text{y}} - 4} \right) + 3\left( {{\text{y}} - 4} \right) + 12 = 0
x(y4)+3(y4)=12\Rightarrow {\text{x}}\left( {{\text{y}} - 4} \right) + 3\left( {{\text{y}} - 4} \right) = - 12
(x+3)(y4)=12\Rightarrow \left( {x + 3} \right)\left( {{\text{y}} - 4} \right) = - 12
Thus (x+3)(y4)=0\left( {x + 3} \right)\left( {{\text{y}} - 4} \right) = 0 is the joint equation of asymptotes.
Therefore, xy4x+3y12=0{\text{xy}} - 4{\text{x}} + 3{\text{y}} - 12 = 0 represents the asymptotes of xy4x+3y=0{\text{xy}} - 4{\text{x}} + 3{\text{y}} = 0.
Now, for finding the value of λ{{\lambda }}. We have to compare the given asymptotic equationxy4x+3yλ=0{\text{xy}} - 4{\text{x}} + 3{\text{y}} - {{\lambda }} = 0 with new asymptotic equation xy4x+3y12=0{\text{xy}} - 4{\text{x}} + 3{\text{y}} - 12 = 0. Then we get the desired solution.
Compare xy4x+3yλ=0{\text{xy}} - 4{\text{x}} + 3{\text{y}} - {{\lambda }} = 0 with xy4x+3y12=0{\text{xy}} - 4{\text{x}} + 3{\text{y}} - 12 = 0.
Thus, we get the value of λ=12{{\lambda }} = 12.

\therefore The correct option is D{\text{D}}.

Note: We have to know that an asymptote to a curve is the tangent to the curve such that the point of contact is at infinity. In particular the asymptote touches the curve at ++ \infty and - \infty.
The equation of the asymptote to the hyperbola x2a2y2b2=1\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}} - \dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}} = 1 .