Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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 a length of major axis 6
C. A hyperbola with eccentricity $\sqrt 5 $
D. A hyperbola with a length of conjugate axis 3

seo-qna
Last updated date: 12th Jul 2024
Total views: 381.6k
Views today: 4.81k
Answer
VerifiedVerified
381.6k+ views
Hint: As we are given two straight lines we find the value of t from (2) and using this in equation (1) we get a equation of a conic and comparing it with the general equation of ellipse and hyperbola we get that it is a hyperbola with ${a^2} = 9$ and ${b^2} = \dfrac{9}{4}$ and now we can find the eccentricity using the formula $\sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} $ and conjugate axis is 2b and see which matches the given options.

Complete step by step solution:
We are given two straight lines
$ \Rightarrow tx - 2y - 3t = 0$ ………(1)
$ \Rightarrow x - 2ty + 3 = 0$……….(2)
From the second equation we can find the value of t
$
   \Rightarrow x + 3 = 2ty \\
   \Rightarrow \dfrac{{x + 3}}{{2y}} = t \\
 $
Using the value of t in equation (1)
$
   \Rightarrow \left( {\dfrac{{x + 3}}{{2y}}} \right)x - 2y - 3\left( {\dfrac{{x + 3}}{{2y}}} \right) = 0 \\
   \Rightarrow \left( {\dfrac{{{x^2} + 3x}}{{2y}}} \right) - 2y - \left( {\dfrac{{3x + 9}}{{2y}}} \right) = 0 \\
   \Rightarrow \dfrac{{{x^2} + 3x - 4{y^2} - 3x - 9}}{{2y}} = 0 \\
   \Rightarrow \dfrac{{{x^2} - 4{y^2} - 9}}{{2y}} = 0 \\
 $
 Cross multiplying and taking the constant to the other side we get
$
   \Rightarrow {x^2} - 4{y^2} - 9 = 0 \\
   \Rightarrow {x^2} - 4{y^2} = 9 \\
 $
In the given options we are given that its either a ellipse or hyperbola
Hence to bring it to that form lets divide throughout by 9
 $
   \Rightarrow \dfrac{{{x^2}}}{9} - \dfrac{{4{y^2}}}{9} = \dfrac{9}{9} \\
   \Rightarrow \dfrac{{{x^2}}}{9} - \dfrac{{4{y^2}}}{9} = 1 \\
 $
We know that the general form of the ellipse and hyperbola are $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ and }}\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ }}$ respectively
Hence from this we get that the locus is a hyperbola
In our option we have a hyperbola with eccentricity $\sqrt 5 $ or with a length of conjugate axis 3
So now lets find the eccentricity of our hyperbola and its conjugate axis
In a hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ }}$the eccentricity is given by $\sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} $ and the conjugate axis is given as 2b
Now let's find the eccentricity
Here ${a^2} = 9$ and ${b^2} = \dfrac{9}{4}$
Using this we get
 $
   \Rightarrow e = \sqrt {1 + \dfrac{{\dfrac{9}{4}}}{9}} \\
   \Rightarrow e = \sqrt {1 + \dfrac{1}{4}} \\
   \Rightarrow e = \sqrt {\dfrac{{4 + 1}}{4}} = \sqrt {\dfrac{5}{4}} = \dfrac{{\sqrt 5 }}{2} \\
 $
Here we get the eccentricity to be $\dfrac{{\sqrt 5 }}{2}$which does not match the given option
So now lets find the conjugate axis
The conjugate axis is 2b
Since ${b^2} = \dfrac{9}{4}$we get$b = \sqrt {\dfrac{9}{4}} = \dfrac{3}{2}$
Hence our conjugate axis is
$ \Rightarrow 2b = 2\left( {\dfrac{3}{2}} \right) = 3$
From this we get that the locus is a hyperbola with conjugate axis 3

Therefore the correct answer is option D.

Note :
A hyperbola is created when the plane intersects both halves of a double cone, creating two curves that look exactly like each other, but open in opposite directions.
The eccentricity of a hyperbola is greater than 1. This indicates that the distance between a point on a conic section the nearest directrix is less than the distance between that point and the focus.