### Video Transcript

Find an equation of the tangent to the curve π₯ equals π‘ squared minus π‘, π¦ equals π‘ squared plus π‘ plus one, at the point zero, three.

So what we have here are a pair of parametric equations. And as it says, weβre gonna try and find the equation of the tangent. So weβve been given a point, but what we also want to find is the π‘ value. So therefore, what we can actually do is use our point to find the π‘ value, because we have π₯ is equal to zero and π¦ is equal to three.

So what we can do is actually take our value of zero and actually substitute it into π₯ equals π‘ squared minus π‘, because what we can say is that as weβve said π₯ is zero, so we get π‘ squared minus π‘ equals zero. And then if we factor this, we take out π‘ as a factor cause π‘ is a factor of both π‘ squared and π‘. So we get π‘ and then in parentheses we have π‘ minus one is equal to zero. So therefore, we can say that π‘ is equal to zero or one. Because if π‘ was equal to zero, then that would give us a result of zero to be zero multiplied by negative one, which would just give us zero. Or if π‘ was equal to one, weβd get one multiplied by one minus one which is zero, which again would give us zero.

Okay great! So we found possible π‘ values using our π₯ value. So now letβs use our π¦ value of three. So if we substitute π¦ is equal to three and two π¦ equals π‘ squared plus π‘ plus one, we get π‘ squared plus π‘ plus one is equal to three. Iβve actually just rewritten it like the first one, this way round, just cause it makes it easier to solve. And then if I subtract three from each side of the equation, I get π‘ squared plus π‘ minus two is equal to zero.

And again, I can actually solve this using factoring. So Iβve actually factored π‘ squared plus π‘ minus two. And when I do that, I get π‘ plus two multiplied by π‘ minus one. And we did that, because we found two numbers or two factors that gave us negative two when multiplied together. So positive two multiplied by negative one gives us negative two. And then if you add two and negative one, you get one, which is the coefficient of our π‘ value.

So therefore our possible solutions are negative two or one, so π‘ is equal to negative two or one. And again thatβs because if we put negative two into the first parentheses, then what we get is negative two plus two, which gives us zero, and therefore would give us a result of zero. And similarly, if we put one into the second parentheses, we get one minus one, which gives us zero. So we get our result of zero. Okay! So now weβve actually got possible values of π‘. So which value of π‘ is actually gonna be the value that we want?

But we can actually see that the shared value is actually one. So therefore, our π‘ value that we wanted to use is actually going to be one. Okay! So great! We now have all the information that we need to actually solve the problem and find the equation of the tangent. Now because actually the curve and the tangent at this point are gonna have the same slope, then therefore what weβre gonna want to do is actually find the slope of our curve at this point. So to do that, what we can actually do is find slope function so dπ¦ dπ₯.

And to find this using parametric equations, what weβre gonna do is use this relationship. And the relationship is that dπ¦ dπ₯ is equal to dπ¦ dπ‘ divided by dπ₯ dπ‘. So what this actually means in practice is that actually our slope function is gonna be equal to the derivative of our π¦ equation divided by the derivative of our π₯ equation. Okay, so letβs get on and find dπ¦ dπ‘ and dπ₯ dπ‘. So if we have π₯ equals π‘ squared minus π‘, then dπ₯ dπ‘ is going to be equal to two π‘ minus one.

So Iβm just gonna very quickly remind you how we actually differentiate it. So weβve got the coefficient of one multiplied by the exponent of two, which gives us our two, and then youβve got π‘ to power of two minus one, which gives us just π‘ to the power of one or π‘ on its own. Okay, great! So now letβs move on to π¦ and find dπ¦ dπ‘. So now if we actually differentiate π‘ squared plus π‘ plus one, weβll get two π‘ plus one. Okay, great! So now weβve got dπ₯ dπ‘ and dπ¦ dπ‘. So now what weβre gonna do is actually find our slope function cause we could find dπ¦ dπ₯ by dividing dπ¦ dπ‘ by dπ₯ dπ‘.

So therefore our slope function is equal to two π‘ plus one over two π‘ minus one. Okay, great! But what we want to do is actually find the value of this. So to actually find the slope, what we need to do is actually substitute our value of π‘ equals one into dπ¦ dπ₯. And when you do that, weβre gonna get two multiplied by one plus one divided by two multiplied by one minus one, which gives us three over one, which is just three. So therefore, weβve found our slope or π value.

So we can say that π or our slope is three. So why is this actually useful? Well itβs useful because what weβre trying to do is actually find the equation of the tangent. So the tangent is actually going to be a straight line. So therefore, weβre gonna use the general form of the equation of a straight line, and that is π¦ is equal to ππ₯ plus π where π, as weβve already discussed, is our slope and π is our π¦-intercept. So therefore if we actually substitute our value of π into π¦ equals ππ₯ plus π, we get the equation π¦ is equal to three π₯ plus π.

Okay, great! But now what we need to do is actually find out π. And how are weβre going to do that? Well we can do that cause we actually know a point on our tangent because that point is zero, three. So therefore, what we can do is actually substitute in zero for our π₯ value and three for our π¦ value to help us find π. So if we substitute π₯ equals zero and π¦ equals three, we get three is equal to three multiplied by zero plus π, so therefore giving us a π value or a π¦ intercept of three. So great! We now have our π, our slope, and our π, our π¦-intercept.

What we can do is just substitute these both back into our formula for the equation, and this will give us the equation of the tangent. So therefore, what we can say is that the equation of the tangent to the curve π₯ equals π‘ squared minus π‘, π¦ equals π‘ squared plus π‘ plus one at the point zero, three is π¦ is equal to three π₯ plus three.