Q65 / 200 · WBJEE 2013 · Mathematics

For the variable, the locus of the point of intersection of the lines $3 t x-2 y+6 t=0$ and $3 x+2 t y-6=0$ is

  • Correct answer
    A
    the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$
  • B
    the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$
  • C
    the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{9}=1$
  • D
    the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$

Solution

Given equation of lines are $$ 3 t \times-2 y+6 t=0 $$ and $\quad 3 x+2 t y-6=0$ On multiplying Eq. (i) by $t$ and then adding in Eq. (ii), we get $\left(3 t^{2}+3\right) x+6 t^{2}-6=0$ $\Rightarrow \quad x=\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)}$ $\Rightarrow \quad x+x t^{2}=2-2 t^{2}$ $\Rightarrow \quad(x+2) t^{2}=(2-x)$ $\Rightarrow \quad t^{2}=\frac{2-x}{2+x}$ On multiplying Eq. (ii) by $t$ and then subtract from Eq. (1). we get $$ \begin{array}{r} \left(-2-2 t^{2}\right) y+6 t+6 t=0 \\ 12 t=2\left(1+t^{2}\right) y \end{array} $$ On squaring both sides, we get $144 t^{2}=4 y^{2}\left(1+t^{2}\right)^{2}$ $\Rightarrow 144\left(\frac{2-x}{2+x}\right)=4 y^{2}\left(1+\frac{2-x}{2+x}\right)^{2}$ [form Eq. (ii)] $\Rightarrow \quad 36\left(\frac{2-x}{2+x}\right)=y^{2}\left(\frac{4}{2+x}\right)^{2}$ $\Rightarrow \quad 36 \frac{(2-x)}{(2+x)}=\frac{16 y^{2}}{(2+x)^{2}}$ $\Rightarrow \quad 36\left(4-x^{2}\right)=16 y^{2}$ $\Rightarrow \quad 9\left(4-x^{2}\right)=4 y^{2}$ $\Rightarrow \quad 36-9 x^{2}=4 y^{2}$ $\Rightarrow \quad 9 x^{2}+4 y^{2}=36$ $\Rightarrow \quad \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ which represents an ellipse

Hence, the correct answer is option: 'A'.

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