Question

The locus of point of intersection of the lines
$\frac{tx}{a}-\frac{y}{b}+t=0,\frac{x}{a}+\frac{ty}{b}-1=0$ is $(t$ is a parameter and $a\ne b)$

• A. parabola
• B. circle
• C. pair of straight lines
• D. ellipse

Answer 1

The correct option is D ellipse

$\frac{tx}{a}-\frac{y}{b}+t=0\cdots \left(1\right)$
$\frac{x}{a}+\frac{ty}{b}-1=0\cdots \left(2\right)$
From equations $\left(1\right)$ and $\left(2\right),$ the point of intersetion is
$(h,k)\equiv (a\left(\frac{1-t^2}{1+t^2}\right),\frac{2bt}{1+t^2})\therefore \frac{h}{a}=\frac{1-t^2}{1+t^2},\frac{k}{b}=\frac{2t}{1+t^2}\Rightarrow \frac{h^2}{a^2}+\frac{k^2}{b^2}=\frac{(t^2-1{)}^2+4t^2}{(1+t^2{)}^2}\Rightarrow \frac{h^2}{a^2}+\frac{k^2}{b^2}=\frac{(t^2+1{)}^2}{(1+t^2{)}^2}=1$

Hence locus will be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Which is an ellipse.