Question

The locus of the point of intersection of the straight lines, \[tx - 2y - 3t = 0\], \[x - 2ty + 3 = 0\left( {t \in R} \right)\], is.
A. An ellipse with eccentricity \[\dfrac{2}{{\sqrt 5 }}\]
B. An ellipse with the length of major axis 6
C. A hyperbola with eccentricity \[\sqrt 5 \]
D. A hyperbola with the length of conjugate axis 3

Answer 1

Hint: Here, we will find the value of \[t\] from the equation \[x - 2ty + 3 = 0\] and then substitute the obtained value in equation \[tx - 2y - 3t = 0\]. Then simply find the eccentricity and conjugate axis.

Complete step by step answer:

We are given that equation of lines are
\[tx - 2y - 3t = 0{\text{ .......eq.(1)}}\]
\[x - 2ty + 3 = 0{\text{ .......eq.(2)}}\]

Subtracting the equation (2) by \[x + 3\] on both sides, we get

\[
   \Rightarrow x - 2ty + 3 - \left( {x + 3} \right) = 0 - \left( {x + 3} \right) \\
   \Rightarrow x - 2ty + 3 - x - 3 = - x - 3 \\
   \Rightarrow - 2ty = - x - 3 \\
 \]

Dividing the above equation by \[ - 2y\] on both sides, we get

\[
   \Rightarrow \dfrac{{ - 2ty}}{{ - 2y}} = \dfrac{{ - x - 3}}{{ - 2y}} \\
   \Rightarrow t = \dfrac{{x + 3}}{{2y}} \\
 \]

Substituting this value in the equation (1), we get

\[
   \Rightarrow \left( {x - 3} \right) \times \dfrac{{x + 3}}{{2y}} - 2y = 0 \\
   \Rightarrow \dfrac{{{x^2} - 9}}{{2y}} - 2y = 0 \\
 \]

Multiplying the above equation by \[2y\] on sides, we get

\[
   \Rightarrow 2y\left( {\dfrac{{{x^2} - 9}}{{2y}} - 2y} \right) = 0 \\
   \Rightarrow {x^2} - 9 - 4{y^2} = 0 \\
 \]

Adding the above equation by 9 on both sides, we get

\[
   \Rightarrow {x^2} - 9 - 4{y^2} + 9 = 0 + 9 \\
   \Rightarrow {x^2} - 4{y^2} = 9 \\
   \Rightarrow \dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{{\dfrac{9}{4}}} = 1{\text{ ......eq.(3)}} \\
 \]
We know that the general equation of a hyperbola is \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\], where \[a\] is line segment for the \[x\]–axis and \[b\] is the line segment for the \[y\]–axis.
This implies that the above equation is a hyperbola by comparing the general equation of a hyperbola.

We also know that the eccentricity of a hyperbola is always greater than 1 and can be calculated using the formula, \[\sqrt {\dfrac{{{a^2} + {b^2}}}{{{a^2}}}} \].

Replacing 3 for \[a\] and \[\dfrac{3}{2}\] for \[b\] in the above formula of eccentricity, we get

\[
   \Rightarrow \sqrt {\dfrac{{{3^2} + {{\left( {\dfrac{3}{2}} \right)}^2}}}{{{3^2}}}} \\
   \Rightarrow \sqrt {\dfrac{{9 + \dfrac{9}{4}}}{9}} \\
   \Rightarrow \sqrt {\dfrac{{\dfrac{{36 + 9}}{4}}}{9}} \\
   \Rightarrow \sqrt {\dfrac{{45}}{{4 \times 9}}} \\
   \Rightarrow \dfrac{{\sqrt 5 }}{2} \\
 \]

We know that the semi conjugate axis is the length of the line segment with respect to \[y\]–axis from equation (3), we get
\[ \Rightarrow \sqrt {\dfrac{9}{4}} = \dfrac{3}{2}\]

Therefore, the length of the conjugate axis is 3.
Hence, option D is correct.

Note: In solving these types of questions, students find the value from one equation and substitute it in the other equation to find the equation of the hyperbola. The 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 by knowing the properties to the general equation of hyperbola \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\], the solution will be very simple.