Question

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

• 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 $\frac{x^2}{9}-\frac{y^2}{4}=1$

Answer 1

The correct option is A the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$

Given equation of lines are
$3tx-2y+6t=\mathrm{0......}\left(i\right)$
and

$3x+2ty-6=\mathrm{0.....}\left(ii\right)$

On multiplying Eq $\left(i\right)$ by $t$ and then adding in eq.$\left(ii\right)$, we get
$(3t^2+3)x+6t^2-6=0$
$\Rightarrow$ $x=\frac{2(-1t^2)}{(1+t^2)}$
$\Rightarrow$ $x+x{t}^2=2-2t^2$
$\Rightarrow$ $(x+2)t^2=(2-x)$
$\Rightarrow$ $t^2=\frac{2-x}{2+x}.....\left(iii\right)$

On multiplying eq $\left(ii\right)$ by $t$ and then subtract from eq $\left(i\right)$ we get
$(-2-2t^2)y+6t+6t=0$
$12t=2(1+t^2)y$

On squaring both sides, we get
$144t^2=4y^2{(1+t^2)}^2$
$\Rightarrow$ $144\left(\frac{2-x}{2+x}\right)=4y^2{(1+\frac{2-x}{2+x})}^2$ (from Eq $\left(iii\right)$]
$\Rightarrow$ $36\left(\frac{2-x}{2+x}\right)=y^2{\left(\frac{4}{2+x}\right)}^2$

$\Rightarrow$ $36\frac{2-x}{2+x}=\frac{16y^2}{{(2+x)}^2}$

$\Rightarrow$ $36(4-x^2)=16y^2$

$\Rightarrow$ $9(4-x^2)=4y^2$

$\Rightarrow$ $36-9x^2=4y^2$

$\Rightarrow$ $9x^2+4y^2=36$

$\Rightarrow$ $\frac{x^2}{4}+\frac{y^2}{9}=1$ which represents an ellipse