Question Video: Plotting a Locus When Given an Equation to Represent Its Argument | Nagwa Question Video: Plotting a Locus When Given an Equation to Represent Its Argument | Nagwa

Question Video: Plotting a Locus When Given an Equation to Represent Its Argument Mathematics

The locus of π§ satisfies arg ((π§ β 3π)/(π§ β 5π)) = 2π/3. Sketch this locus on an Argand diagram and find its Cartesian equation.

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Video Transcript

The locus of π§ satisfies the argument of π§ minus three π over π§ minus five π is equal to two π by three. Sketch this locus on an Argand diagram and find its Cartesian equation.

Remember, the locus of a point π§ such that the argument of π§ minus π§ one over π§ minus π§ two is π is the arc of a circle which subtends an angle of π between the points represented by π§ one and π§ two. Itβs traced in a counterclockwise direction from π§ one and π§ two, but the endpoints arenβt part of the locus. Now, we know that if π is greater than π by two radians, the locus is a minor arc. Similarly, if itβs greater than π by two, itβs a major arc. And if itβs equal to π by two, itβs a semicircle.

Here, we can say that π§ one must be three π, π§ two must be five π, and π is two π by three radians. So that is greater than π by two, and itβs a minor arc. Now, to plot π§ one and π§ two on the Argand plane, we plot the points with Cartesian coordinates zero, three and zero, five, respectively. But here we have a bit of a problem. How do we know where the minor arc sits? We know itβs the arc of a circle. But without knowing the center of the circle, we canβt use this information to find the arc. We do, however, know that the locus is drawn in a counterclockwise direction from π§ one to π§ two. And for the arc from zero, three to zero, five to be a minor arc when drawn in this direction, we need to draw the locus shown. And we can add two π by three radians if necessary.

The next part of this question asks us to find the Cartesian equation of the locus. Now, in some scenarios, we can find the Cartesian equation by finding the center and radius of the circle. In this case though, weβre going to use an algebraic approach. Weβre going to substitute π§ equals π₯ plus π¦π into the complex equation. So we get the argument of π₯ plus π¦π minus three π over π₯ plus π¦π minus five π equals two π over three. Letβs rewrite this a little bit. And then weβre going to perform the complex division π₯ plus π times π¦ minus three divided by π₯ plus π times π¦ minus five.

To do this, we multiply the numerator and denominator of our fraction by the conjugate of the denominator. Remember, to find the conjugate of our denominator, we change the sign of the imaginary part. So weβre going to multiply the numerator and denominator of our fraction by π₯ minus π times π¦ minus five. Distributing the parentheses on our numerator and remembering of course that π squared is negative one, we get π₯ squared minus π₯π times π¦ minus five plus π₯π times π¦ minus three plus π¦ minus three times π¦ minus five. And on our denominator, we simply get π₯ squared plus π¦ minus five squared.

Weβre now going to split this into the real and imaginary parts. And we see that the real part is π₯ squared plus π¦ minus three times π¦ minus five over π₯ squared plus π¦ minus five squared. And the imaginary part is negative π₯ times π¦ minus five plus π₯ times π¦ minus three, which Iβve written as π₯ times π¦ minus three minus π₯ times π¦ minus five all over π₯ squared plus π¦ minus five squared.

Now, in fact, distributing the parentheses, and we get π₯ squared plus π¦ squared minus eight π¦ plus 15 as the numerator of our first part and two π₯ as the numerator of our imaginary part. Now, we know that the argument of this complex number is equal to two π by three. And for a complex number of the form π plus ππ, we know that tan of π, where π is the argument, is equal to π over π. This means that tan of two π over three will be equal to π, which is the imaginary part, divided by π, which is the real part of our complex number. And when we divide the imaginary part by the real part, the denominators cancel. So we find that tan of two π over three equals two π₯ over π₯ squared plus π¦ squared minus eight π¦ plus 15. tan of two π over three, though, is negative root three.

Letβs clear some space and rearrange our equation. To do so, weβre going to multiply both sides by the denominator of our fraction and then divide through by negative root three. And when we do, we find that π₯ squared plus π¦ squared minus eight π¦ plus 15 is equal to two π₯ divided by negative root three, which we can write as negative two root three over three π₯. We add two root three over three π₯ to both sides, and now weβre going to complete the square. Completing the square for the expression π₯ squared plus two root three over three π₯ gives us π₯ plus root three over three squared minus a third.

Similarly, completing the square for π¦ squared minus eight π¦, and we get π¦ minus four squared minus 16. And of course, we add 15 and set this equal to zero. Negative one-third minus 16 plus 15 is negative four-thirds. So we add four-thirds to both sides. And we get the equation π₯ plus root three over three squared plus π¦ minus four all squared equals four-thirds.

We can now see that this is the equation of a circle as we were expecting. Now, we do need to be a little bit careful. We need to state a restriction on π₯ and π¦ to ensure that the points lie on the locus. And of course, thatβs the minor arc we sketched earlier. And so weβre only going to consider the points where π₯ is greater than zero. And so weβve sketched the locus on an Argand diagram and found its Cartesian equation to be π₯ plus root three over three squared plus π¦ minus four squared equals four-third for values of π₯ greater than zero.

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