Question

Find the locus of the point of intersection of two straight lines $\dfrac{tx}{a}-\dfrac{y}{b}+t=0$and$\dfrac{x}{a}+\dfrac{ty}{b}-1=0$.

Answer 1

Hint: Assume a point and its coordinates, and make them satisfy the two equations of lines given in the question. Then, eliminate $t$ from them, using algebra, and get an equation only in terms of the $x$ and $y$ coordinates of the point of intersection you originally assumed.

We have been given two lines in the equation, let’s start working towards simplifying them first.
Given that,
$\dfrac{tx}{a}-\dfrac{y}{b}+t=0$ ……………. (1)
$\Rightarrow btx-ay=-abt$ ……………… (A)
$\dfrac{x}{a}+\dfrac{ty}{b}-1=0$ ……………… (2)
$\Rightarrow bx+aty=ab$ ……………… (B)
Thus, we have successfully simplified both the equations to represent something like the general equation of a line, which is $y=mx+c$.
Let the given curves intersect each other at a point$P(h,k)$.
Therefore $P(h,k)$will satisfy both the curves. Let’s put the value of $P(h,k)$ in both, equation (A) and equation (B). Doing so, we get :
$bth-ak=-abt$ ……………. (3)
$bh+atk=ab$ …………….. (4)
Multiply the equation (3) with $t$ and add with the equation (4).
Multiplying (3) with $t$, we will get the equation $b{{t}^{2}}h-akt=-ab{{t}^{2}}$
Now, let’s add the new equation we got after multiplying (3), to (4). Doing so, we get :
$\left( b{{t}^{2}}h-akt=-ab{{t}^{2}} \right)$
$bh+atk=ab$
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$\Rightarrow bh+b{{t}^{2}}h=ab-ab{{t}^{2}}$
$\Rightarrow bh\left( 1+{{t}^{2}} \right)=ab\left( 1-{{t}^{2}} \right)$
$\Rightarrow h=\dfrac{a(1-{{t}^{2}})}{(1+{{t}^{2}})}$
$\Rightarrow \dfrac{h}{a}=\dfrac{\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)}$
Squaring both sides, we get :
$\Rightarrow {{\left( \dfrac{h}{a} \right)}^{2}}={{\left( \dfrac{\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)} \right)}^{2}}$
$\Rightarrow \dfrac{{{h}^{2}}}{{{a}^{2}}}=\dfrac{{{\left( 1-{{t}^{2}} \right)}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}$ ……….. (5)
Put the value$h=\dfrac{a\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)}$ in equation (4), we get
$b\left( \dfrac{a\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)} \right)+atk=ab$
$\Rightarrow tk=b-\dfrac{b\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)}$
$\Rightarrow k=\dfrac{b}{t}\left( \dfrac{1+{{t}^{2}}-1+{{t}^{2}}}{1+{{t}^{2}}} \right)$
$\Rightarrow k=\dfrac{2bt}{1+{{t}^{2}}}$
$\Rightarrow \dfrac{k}{b}=\dfrac{2t}{1+{{t}^{2}}}$
Squaring both sides, we get
$\Rightarrow {{\left( \dfrac{k}{b} \right)}^{2}}={{\left( \dfrac{2t}{1+{{t}^{2}}} \right)}^{2}}$
$\Rightarrow \dfrac{{{k}^{2}}}{{{b}^{2}}}=\dfrac{4{{t}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}$ ……………. (6)
On adding equation (5) and equation (6), we get :
$\dfrac{{{h}^{2}}}{{{a}^{2}}}=\dfrac{{{\left( 1-{{t}^{2}} \right)}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}$
$\dfrac{{{k}^{2}}}{{{b}^{2}}}=\dfrac{4{{t}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}$
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$\dfrac{{{h}^{2}}}{{{a}^{2}}}+\dfrac{{{k}^{2}}}{{{b}^{2}}}=\dfrac{{{\left( 1-{{t}^{2}} \right)}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}+\dfrac{4{{t}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}$
$\Rightarrow \dfrac{{{h}^{2}}}{{{a}^{2}}}+\dfrac{{{k}^{2}}}{{{b}^{2}}}=\dfrac{1+{{t}^{4}}-2{{t}^{2}}+4{{t}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}=\dfrac{1+{{t}^{4}}+2{{t}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}$

$\Rightarrow \dfrac{{{h}^{2}}}{{{a}^{2}}}+\dfrac{{{k}^{2}}}{{{b}^{2}}}=1$ This is the locus of point of intersection in $(h,k)$.
Now, to finally find our locus in terms of $x$ and $y$, all we have to do is this :
Replace $(h,k)\to (x,y)$. Doing so, we get :
$\Rightarrow \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
Therefore, we can say that the locus of points of intersection of a given curve is an ellipse.

Note: The given equations are in parametric from with parameter $t$. For different values of $t$, we get a different equation of straight line.