### 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 letβs 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. Thatβs 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 letβs 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 wouldnβt 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. Itβs a standard hyperbola, centred
at the origin with vertices at plus or minus π, zero and covertices at zero, plus
or minus π.

Letβs 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, thereβs only one graph
that satisfies this criteria. Itβs 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.