# The condition that the line \[lx+my+n=0\] may be a tangent to the rectangular hyperbola \[xy={{c}^{2}}\] is

a)\[{{a}^{2}}{{l}^{2}}+{{b}^{2}}{{m}^{2}}={{n}^{2}}\]

b)\[a{{m}^{2}}+nl\]

c)\[{{a}^{2}}{{l}^{2}}-{{b}^{2}}{{m}^{2}}={{n}^{2}}\]

d)\[4{{c}^{2}}lm={{n}^{2}}\]

Last updated date: 23rd Mar 2023

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Answer

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303.9k+ views

Hint: To solve the question, we have to apply the properties of tangent which states that tangent meets the rectangular hyperbola at only one real point. Thus, the line and the rectangular hyperbola have a common point. To calculate the point apply the properties and formulae of quadratic equations.

Complete step-by-step answer:

Let the point \[({{x}_{1}},{{y}_{1}})\] be the point of tangent.

This implies that the point \[({{x}_{1}},{{y}_{1}})\] lie both on the given line and on the rectangular hyperbola.

Thus, the point \[({{x}_{1}},{{y}_{1}})\] will satisfy the equations of rectangular hyperbola and the given line.

Now by substituting the point \[({{x}_{1}},{{y}_{1}})\] in the given equations, we get

\[l{{x}_{1}}+m{{y}_{1}}+n=0\] ….. (1)

\[{{x}_{1}}{{y}_{1}}={{c}^{2}}\]

\[{{x}_{1}}=\dfrac{{{c}^{2}}}{{{y}_{1}}}\]

By substituting the equation (1) in the above equation, we get

\[l\left( \dfrac{{{c}^{2}}}{{{y}_{1}}} \right)+m{{y}_{1}}+n=0\]

\[\dfrac{l{{c}^{2}}+{{y}_{1}}\left( m{{y}_{1}}+n \right)}{{{y}_{1}}}=0\]

\[l{{c}^{2}}+{{y}_{1}}\left( m{{y}_{1}}+n \right)=0\]

\[l{{c}^{2}}+my_{1}^{2}+n{{y}_{1}}=0\]

\[my_{1}^{2}+n{{y}_{1}}+l{{c}^{2}}=0\] …… (2)

The obtained quadratic equation has real roots. Thus, the discriminant of the quadratic equations is equal to 0.

We know that the formula of the discriminant of the general quadratic equation \[a{{x}^{2}}+bx+c=0\] is equal to \[{{b}^{2}}-4ac\]

\[\Rightarrow {{b}^{2}}-4ac=0\]

On comparing with equation (2), we get

\[a=m,b=n,c=l{{c}^{2}}\]

By substituting the values in the discriminant of quadratic equation formula, we get

\[{{n}^{2}}-4m(l{{c}^{2}})=0\]

\[4{{c}^{2}}lm={{n}^{2}}\]

Thus, the condition that the line may be a tangent to the rectangular hyperbola is \[4{{c}^{2}}lm={{n}^{2}}\]

Hence, option (d) is the right choice.

Note: The problem of mistake can be not analysing that the tangent meets the rectangular hyperbola at only one point. The other possibility of mistake is not being able to apply the properties and formulae of quadratic equations to solve. The alternative way of solving the question is option elimination method since only option contains the constant c, the other options can be ruled. The information about constants a and b is not mentioned in the question which makes them irrelevant.

Complete step-by-step answer:

Let the point \[({{x}_{1}},{{y}_{1}})\] be the point of tangent.

This implies that the point \[({{x}_{1}},{{y}_{1}})\] lie both on the given line and on the rectangular hyperbola.

Thus, the point \[({{x}_{1}},{{y}_{1}})\] will satisfy the equations of rectangular hyperbola and the given line.

Now by substituting the point \[({{x}_{1}},{{y}_{1}})\] in the given equations, we get

\[l{{x}_{1}}+m{{y}_{1}}+n=0\] ….. (1)

\[{{x}_{1}}{{y}_{1}}={{c}^{2}}\]

\[{{x}_{1}}=\dfrac{{{c}^{2}}}{{{y}_{1}}}\]

By substituting the equation (1) in the above equation, we get

\[l\left( \dfrac{{{c}^{2}}}{{{y}_{1}}} \right)+m{{y}_{1}}+n=0\]

\[\dfrac{l{{c}^{2}}+{{y}_{1}}\left( m{{y}_{1}}+n \right)}{{{y}_{1}}}=0\]

\[l{{c}^{2}}+{{y}_{1}}\left( m{{y}_{1}}+n \right)=0\]

\[l{{c}^{2}}+my_{1}^{2}+n{{y}_{1}}=0\]

\[my_{1}^{2}+n{{y}_{1}}+l{{c}^{2}}=0\] …… (2)

The obtained quadratic equation has real roots. Thus, the discriminant of the quadratic equations is equal to 0.

We know that the formula of the discriminant of the general quadratic equation \[a{{x}^{2}}+bx+c=0\] is equal to \[{{b}^{2}}-4ac\]

\[\Rightarrow {{b}^{2}}-4ac=0\]

On comparing with equation (2), we get

\[a=m,b=n,c=l{{c}^{2}}\]

By substituting the values in the discriminant of quadratic equation formula, we get

\[{{n}^{2}}-4m(l{{c}^{2}})=0\]

\[4{{c}^{2}}lm={{n}^{2}}\]

Thus, the condition that the line may be a tangent to the rectangular hyperbola is \[4{{c}^{2}}lm={{n}^{2}}\]

Hence, option (d) is the right choice.

Note: The problem of mistake can be not analysing that the tangent meets the rectangular hyperbola at only one point. The other possibility of mistake is not being able to apply the properties and formulae of quadratic equations to solve. The alternative way of solving the question is option elimination method since only option contains the constant c, the other options can be ruled. The information about constants a and b is not mentioned in the question which makes them irrelevant.

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