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 ${\displaystyle \frac{2}{\sqrt{5}}}$
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

Answer 1

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 $b^2={\displaystyle \frac{9}{4}}$ and now we can find the eccentricity using the formula $\sqrt{1+{\displaystyle \frac{b^2}{a^2}}}$ 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
$$\begin{gathered} \Rightarrow x+3=2ty \\ \Rightarrow {\displaystyle \frac{x+3}{2y}}=t \end{gathered}$$
Using the value of t in equation (1)
$$\begin{gathered} \Rightarrow \left({\displaystyle \frac{x+3}{2y}}\right)x-2y-3\left({\displaystyle \frac{x+3}{2y}}\right)=0 \\ \Rightarrow \left({\displaystyle \frac{x^2+3x}{2y}}\right)-2y-\left({\displaystyle \frac{3x+9}{2y}}\right)=0 \\ \Rightarrow {\displaystyle \frac{x^2+3x-4y^2-3x-9}{2y}}=0 \\ \Rightarrow {\displaystyle \frac{x^2-4y^2-9}{2y}}=0 \end{gathered}$$
 Cross multiplying and taking the constant to the other side we get
$$\begin{gathered} \Rightarrow x^2-4y^2-9=0 \\ \Rightarrow x^2-4y^2=9 \end{gathered}$$
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
 $$\begin{gathered} \Rightarrow {\displaystyle \frac{x^2}{9}}-{\displaystyle \frac{4y^2}{9}}={\displaystyle \frac{9}{9}} \\ \Rightarrow {\displaystyle \frac{x^2}{9}}-{\displaystyle \frac{4y^2}{9}}=1 \end{gathered}$$
We know that the general form of the ellipse and hyperbola are ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1\text{ and }{\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1$ 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 ${\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1$the eccentricity is given by $\sqrt{1+{\displaystyle \frac{b^2}{a^2}}}$ and the conjugate axis is given as 2b
Now let's find the eccentricity
Here $a^2=9$ and $b^2={\displaystyle \frac{9}{4}}$
Using this we get
 $$\begin{gathered} \Rightarrow e=\sqrt{1+{\displaystyle \frac{{\displaystyle \frac{9}{4}}}{9}}} \\ \Rightarrow e=\sqrt{1+{\displaystyle \frac{1}{4}}} \\ \Rightarrow e=\sqrt{{\displaystyle \frac{4+1}{4}}}=\sqrt{{\displaystyle \frac{5}{4}}}={\displaystyle \frac{\sqrt{5}}{2}} \end{gathered}$$
Here we get the eccentricity to be ${\displaystyle \frac{\sqrt{5}}{2}}$which does not match the given option
So now lets find the conjugate axis
The conjugate axis is 2b
Since $b^2={\displaystyle \frac{9}{4}}$we get$b=\sqrt{{\displaystyle \frac{9}{4}}}={\displaystyle \frac{3}{2}}$
Hence our conjugate axis is
$\Rightarrow 2b=2\left({\displaystyle \frac{3}{2}}\right)=3$
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.