Question

 The line \[lx + my + n = 0\] will be tangent to the hyperbola \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\], if
A.\[{a^2}{l^2} - {b^2}{m^2} = {n^2}\]
B.\[{a^2}{l^2} + {b^2}{m^2} = {n^2}\]
C.\[a{m^2} - {b^2}{n^2} = {a^2}{l^2}\]
D.None of these

Answer 1

Hint: In this question, we will use the property that if \[y = Mx + C\] is tangent to hyperbola \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\], then \[{C^2} = {a^2}{M^2} - {b^2}\] and rewrite equation of the line \[lx + my + n = 0\] to compare with \[y = Mx + C\] to find the value of \[M\] and \[C\]. Then we will substitute the above values of \[M\] and \[C\] in the equation.

Complete step by step answer:
We are given that the line \[lx + my + n = 0\] will be tangent to the hyperbola \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\].

We know that if \[y = Mx + C\] is tangent to hyperbola \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\], then \[{C^2} = {a^2}{M^2} - {b^2}\].

Rewriting the equation of the line \[lx + my + n = 0\] by dividing both sides with \[m\], we get

\[
   \Rightarrow \dfrac{{lx + my + n}}{m} = 0 \\
   \Rightarrow \dfrac{l}{m}x + \dfrac{m}{m}y + \dfrac{n}{m} = 0 \\
   \Rightarrow \dfrac{l}{m}x + y + \dfrac{n}{m} = 0 \\
 \]

Subtracting the above equation by \[\dfrac{l}{m}x + \dfrac{n}{m}\] on each side, we get

\[
   \Rightarrow y + \dfrac{l}{m}x + \dfrac{n}{m} - \dfrac{l}{m}x - \dfrac{n}{m} = - \dfrac{l}{m}x - \dfrac{n}{m} \\
   \Rightarrow y = - \dfrac{l}{m}x - \dfrac{n}{m} \\
 \]


Comparing the above equation with \[y = Mx + C\] to find the value of \[M\] and \[C\], we get

\[ \Rightarrow M = - \dfrac{1}{m}\]
\[ \Rightarrow C = - \dfrac{n}{m}\]

Now, substituting the above values of \[M\] and \[C\] in the equation \[{C^2} = {a^2}{M^2} - {b^2}\], we get

\[
   \Rightarrow {\left( {\dfrac{n}{m}} \right)^2} = {a^2}{\left( {\dfrac{l}{m}} \right)^2} - {b^2} \\
   \Rightarrow \dfrac{{{n^2}}}{{{m^2}}} = {a^2}\dfrac{{{l^2}}}{{{m^2}}} - {b^2} \\
 \]

Multiplying the above equation by \[{m^2}\] on both sides, we get

\[
   \Rightarrow {m^2}\left( {\dfrac{{{n^2}}}{{{m^2}}}} \right) = {m^2}\left( {{a^2}\dfrac{{{l^2}}}{{{m^2}}} - {b^2}} \right) \\
   \Rightarrow {n^2} = {a^2}{l^2} - {b^2}{m^2} \\
   \Rightarrow {a^2}{l^2} - {b^2}{m^2} = {n^2} \\
 \]

Hence, option A is correct.

Note: In such types of problems, students assume that the tangent meets the hyperbola at only one point, which is wrong. The other possibility of a mistake is not being able to apply the formula and properties of quadratic equations to solve. The key step to solve this problem is knowing the property that if \[y = Mx + C\] is tangent to hyperbola \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\], then \[{C^2} = {a^2}{M^2} - {b^2}\], the solution will be very simple.