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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 :

\[\left( b{{t}^{2}}h-akt=-ab{{t}^{2}} \right)\]

\[bh+atk=ab\]

----------------

\[\Rightarrow bh+b{{t}^{2}}h=ab-ab{{t}^{2}}\]

\[\Rightarrow bh\left( 1+{{t}^{2}} \right)=ab\left( 1-{{t}^{2}} \right)\]

$\Rightarrow h=\dfrac{a(1-{{t}^{2}})}{(1+{{t}^{2}})}$

\[\Rightarrow \dfrac{h}{a}=\dfrac{\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)}\]

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 :

\[\dfrac{{{h}^{2}}}{{{a}^{2}}}=\dfrac{{{\left( 1-{{t}^{2}} \right)}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}\]

\[\dfrac{{{k}^{2}}}{{{b}^{2}}}=\dfrac{4{{t}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}\]

----------------------

\[\dfrac{{{h}^{2}}}{{{a}^{2}}}+\dfrac{{{k}^{2}}}{{{b}^{2}}}=\dfrac{{{\left( 1-{{t}^{2}} \right)}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}+\dfrac{4{{t}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}\]

\[\Rightarrow \dfrac{{{h}^{2}}}{{{a}^{2}}}+\dfrac{{{k}^{2}}}{{{b}^{2}}}=\dfrac{1+{{t}^{4}}-2{{t}^{2}}+4{{t}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}=\dfrac{1+{{t}^{4}}+2{{t}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}\]

\[\Rightarrow \dfrac{{{h}^{2}}}{{{a}^{2}}}+\dfrac{{{k}^{2}}}{{{b}^{2}}}=1\] 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.

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 :

\[\left( b{{t}^{2}}h-akt=-ab{{t}^{2}} \right)\]

\[bh+atk=ab\]

----------------

\[\Rightarrow bh+b{{t}^{2}}h=ab-ab{{t}^{2}}\]

\[\Rightarrow bh\left( 1+{{t}^{2}} \right)=ab\left( 1-{{t}^{2}} \right)\]

$\Rightarrow h=\dfrac{a(1-{{t}^{2}})}{(1+{{t}^{2}})}$

\[\Rightarrow \dfrac{h}{a}=\dfrac{\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)}\]

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 :

\[\dfrac{{{h}^{2}}}{{{a}^{2}}}=\dfrac{{{\left( 1-{{t}^{2}} \right)}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}\]

\[\dfrac{{{k}^{2}}}{{{b}^{2}}}=\dfrac{4{{t}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}\]

----------------------

\[\dfrac{{{h}^{2}}}{{{a}^{2}}}+\dfrac{{{k}^{2}}}{{{b}^{2}}}=\dfrac{{{\left( 1-{{t}^{2}} \right)}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}+\dfrac{4{{t}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}\]

\[\Rightarrow \dfrac{{{h}^{2}}}{{{a}^{2}}}+\dfrac{{{k}^{2}}}{{{b}^{2}}}=\dfrac{1+{{t}^{4}}-2{{t}^{2}}+4{{t}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}=\dfrac{1+{{t}^{4}}+2{{t}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}\]

\[\Rightarrow \dfrac{{{h}^{2}}}{{{a}^{2}}}+\dfrac{{{k}^{2}}}{{{b}^{2}}}=1\] 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.

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