Question

The equation :-

$x^2y^2 - 2xy^2 - 3y^2 - 4x^2y + 8xy + 12y = 0.$ reperesents :-

Answer 1

Topic :-

Locus

Given :-

x²y² - 2xy² - 3y² - 4x²y + 8xy + 12y = 0

To Find :-

Locus of the given equation.

Solution :-

x²y² - 2xy² - 3y² - 4x²y + 8xy + 12y = 0

Taking y² common from few terms,

y²(x² - 2x - 3) - 4x²y + 8xy + 12y = 0

Taking - 4y common from remaining terms,

y²(x² - 2x - 3) - 4y(x² - 2x - 3) = 0

Taking ( x² - 2x - 3 ) common,

(x² - 2x - 3)(y² - 4y) = 0

Splitting (x² - 2x - 3) to factorise it,

(x² - 3x + x - 3)(y² - 4y) = 0

(x(x - 3) + 1(x - 3))(y² - 4y) = 0

Taking (x - 3) common,

(x + 1)(x - 3)(y² - 4y) = 0

Taking y common from (y² - 4y),

(x + 1)(x - 3)y(y-4) = 0

Now,

Any factor can be equal to Zero.

So,

x + 1 = 0

x = -1

A line passing through x = -1 and parallel to Y-axis.

x - 3 = 0

x = 3

A line passing through x = 3 and parallel to Y-axis.

y = 0

It represents X-axis.

y - 4 = 0

y = 4

A line passing through y = 4 and parallel to X-axis.

So, we get four lines.

On plotting these lines on a graph paper, we get to know that these lines form a square of 4 units.

In the graph,

Red line represents x = 3.

Blue line represents x = -1.

Green line represents y = 0.

Violet line represents y = 4.

Answer :-

So, locus of the given equation represents 4 lines which forms a square of side 4 units.

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\: {x}^{2} {y}^{2} - 2 {xy}^{2} - {3y}^{2} - {4yx}^{2} + 8xy + 12y = 0$

can be rewritten as

$\rm :\longmapsto\:( {x}^{2}{y}^{2} - {2xy}^{2} - {3y}^{2}) - ( {4yx}^{2} - 8xy - 12y) = 0$

$\rm :\longmapsto\: {y}^{2}( {x}^{2} - 2x - 3) - 4y( {x}^{2} - 2x - 3) = 0$

$\rm :\longmapsto\:( {x}^{2} - 2x - 3)( {y}^{2} - 4y) = 0$

$\rm :\longmapsto\:( {x}^{2} - 3x + x - 3)y(y - 4) = 0$

$\rm :\longmapsto\:\bigg(x(x - 3) + 1(x - 3) \bigg) y(y - 4( = 0$

$\rm :\longmapsto\:(x - 3)(x + 1)y(y - 4) = 0$

$\rm :\implies\:x = 3 \: or \: x = - 1 \: or \: y = 0 \: or \: y = 4$

If we plot these lines, on graph,

(please see the attachment)

We found that

$\rm :\longmapsto\: {x}^{2} {y}^{2} - 2 {xy}^{2} - {3y}^{2} - {4yx}^{2} + 8xy + 12y = 0$

$\sf \: represents \: a \: square \: of \: length \: 4 \: units.$

Thus,

$\rm :\longmapsto\: {x}^{2} {y}^{2} - 2 {xy}^{2} - {3y}^{2} - {4yx}^{2} + 8xy + 12y = 0$

$\begin{gathered}\begin{gathered}\bf\: represents \: 4 \: lines-\begin{cases} &\sf{x = - 1} \\ &\sf{x = 3}\\ &\sf{y = 0}\\ &\sf{y = 4} \end{cases}\end{gathered}\end{gathered}$