Question Video: Multiplying Imaginary Numbers | Nagwa Question Video: Multiplying Imaginary Numbers | Nagwa

# Question Video: Multiplying Imaginary Numbers Mathematics • Higher Education

Consider the rectangular equation π₯Β² β π¦Β² = 25. Convert the given equation to polar form. Which of the following is the sketch of the equation?

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### Video Transcript

Consider the rectangular equation π₯ squared minus π¦ squared equals 25. Convert the given equation to polar form. The second part of this question asks us, which of the following is the sketch of the equation?

So we begin by recalling that we can convert from polar to Cartesian coordinates using the formulae π₯ equals π cos π and π¦ equals π sin π. Our equation contains π₯ squared and π¦ squared. So letβs square each of these formulae. And when we do, we find that π₯ squared equals π squared cos squared π and π¦ squared equals π squared sin squared π. We know the difference of these is equal to 25. Thatβs our rectangular equation. So π squared cos squared π minus π squared sin squared π equals 25.

We can then factor π squared. So π squared times cos squared π minus sin squared π equals 25. But we know that cos two π is equal to cos squared π minus sin squared π. So letβs replace cos squared π minus sin squared π with cos two π. That tells us that π squared times cos two π equals 25. And we can then divide both sides of this equation by cos two π. Now of course, one over cos π is sec π. So we find that π squared is equal to 25 sec two π.

For part two, we need to identify which of the following is a sketch of the equation. Now we wouldnβt be particularly easy to sketch the graph of π squared equals 25 sec two π. But we do know the general form of the graph whose equation is π₯ over π all squared minus π¦ over π all squared equals one. Itβs a standard hyperbola, centred at the origin with vertices at plus or minus π, zero and covertices at zero, plus or minus π.

Letβs rearrange our equation to equate it to one. To do that, we divide everything through by 25. And since 25 is five squared, we can write this as π₯ over five all squared minus π¦ over five all squared equals one. We know we have a standard hyperbola with vertices at plus or minus five, zero. And in fact, thereβs only one graph that satisfies this criteria. Itβs graph A. Of course, it is useful to know that if we were struggling, we could even try substituting some values of π₯ or π¦ in and sketching the order pairs.

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