Question Video: Finding the Points on an Implicitly Defined Curve Where the Tangents at These Points Are Perpendicular to a Given Line Mathematics • Third Year of Secondary School

Video Transcript

Determine the points on a curve 𝑥 squared plus 𝑦 squared plus 𝑥 plus eight 𝑦 that equals zero at which the tangent is perpendicular to the line seven 𝑦 plus four 𝑥 plus 𝑐 equals zero.

So the first part of this question we’re actually gonna have a look at is bit where it says the tangent is perpendicular to the line and that line is seven 𝑦 plus four 𝑥 plus 𝑐 equals zero. Okay, so let’s use this first.

So if we have our line, then what we wanna do is we actually want to rearrange it. And we want to rearrange it into the form 𝑦 equals 𝑚𝑥 plus 𝑐 because that way as you can see here 𝑚 is going to be the slope and the plus 𝑐 will be the 𝑦-intercept. And in this case it’s actually the slope that we’re looking for. So if we subtract four 𝑥 and 𝑐, we get seven 𝑦 is equal to negative four 𝑥 minus 𝑐. And then if we divide both sides by seven, we get 𝑦 is equal to negative four 𝑥 minus 𝑐 over seven.

So I’m just gonna rewrite this to make it more clear what the slope’s going to be. Then, we get 𝑦 is equal to negative four over seven 𝑥 minus 𝑐 over seven. So therefore, we can say that the slope is gonna be our 𝑚. So it’ll be negative four over seven.

Okay, so we found the slope. But why is this useful? Well, it’s actually this word here that makes this useful “perpendicular” because actually what we’re looking for is we’re looking for the points on the curve where the tangent is perpendicular to our line. So if it’s perpendicular, then therefore the slope is gonna be the negative reciprocal. So therefore, we can say the slope at the points is going to be seven over four. And this is because this is a negative reciprocal of negative four over seven.

Okay, fantastic, now we know what slope we’re looking for. We need to now actually work out the slope function of our curve. So if we have the function 𝑥 squared plus 𝑦 squared plus 𝑥 plus eight 𝑦 is equal to zero, then what we’re gonna do is actually gonna differentiate this to find our slope function. In order to do this, we’re gonna use implicit differentiation.

So the first stage of implicit differentiation, which I wouldn’t normally write up, but I just put here to show actually what’s happening, is we actually differentiate each of our terms with respect to 𝑥 which is gonna give us two 𝑥 plus two 𝑦 𝑑𝑦 𝑑𝑥 plus one plus eight 𝑑𝑦 𝑑𝑥 equals zero. Just to remind us how we got the second and fourth terms, if we have a term that’s in 𝑦 — so it’s a functional term in 𝑦 — and we want to differentiate it with respect to 𝑥, then this is equal to the same functional term differentiated with respect to 𝑦 multiplied by 𝑑𝑦 𝑑𝑥. And this comes from actually adapting the chain rule.

So therefore, if we look at our second term, if we had differentiating 𝑦 squared with respect to 𝑥, then what we can say is this going to be equal to two 𝑦 because if you differentiate 𝑦 squared with respect to 𝑦, you get two 𝑦 because you multiply the coefficient by the exponents — so two by one which gives us two — and then you reduce the exponent by one, so just leaving us with 𝑦. And then, we multiply this by 𝑑𝑦 𝑑𝑥.

Okay, fab, we now know where these have come from, we can move on to the next stage. Well, the next stage is actually to make 𝑑𝑦 𝑑𝑥 the subject because as I said before this is our slope function and this is what we’re looking to find. So therefore, we’re gonna get two 𝑦 𝑑𝑦 𝑑𝑥 plus eight 𝑑𝑦 𝑑𝑥 equals negative two 𝑥 minus one. And now actually as we have 𝑑𝑦 𝑑𝑥 as a factor in both of our terms on the left-hand side, we can actually factor. So therefore, we get 𝑑𝑦 𝑑𝑥 multiplied by two 𝑦 plus eight equals negative two 𝑥 minus one.

So therefore, to finally find our slope function, all we need to do is divide both sides by two 𝑦 plus eight. So therefore, we can say 𝑑𝑦 𝑑𝑥 — so our slope function — is equal to negative two 𝑥 minus one over two 𝑦 plus eight.

Okay, fantastic, so we’ve now got this. What’s the next stage? Well, from early, we actually calculated what the slope would be on the points on our curve where the tangent is perpendicular to the line seven 𝑦 plus four 𝑥 plus 𝑐 equals zero. So what we can do is we can actually substitute in our value for the slope into our slope function and then solve to find the equation of our tangent.

So now, we’ve actually substituted in seven over four for our slope because we know that’s the slope that we want at those points. So we can say that negative two 𝑥 minus one over two 𝑦 plus eight equals seven over four. So therefore, we can say that four multiplied by negative two 𝑥 minus one is equal to seven multiplied by two 𝑦 plus eight which is gonna give us negative eight 𝑥 minus four equals 14𝑦 plus 56, which if we rearrange gives us 14𝑦 plus eight 𝑥 plus 56 plus four equals zero. And then if we collect together all the terms, we have 14𝑦 plus eight 𝑥 plus 60 equals zero.

Then, one more stage, we just divide by two to make it a bit easier to deal with. So now, we have the equation of the tangent at the points that we’re looking for and that is seven 𝑦 plus four 𝑥 plus 30 is equal to zero.

Now, this is the stage where we actually look back at the question because we’ve done lots of things and found lots of different parts here. But what is it the question actually wants? Well, the question wants the points on the curve. So in that case, we need to find out 𝑥- and 𝑦-values. And to do that, we’re gonna set up a simultaneous equation.

So as you can see the equations that I’m going to solve simultaneously are actually the function of our curve, so 𝑥 squared plus 𝑦 squared plus 𝑥 plus eight 𝑦 equals zero, and the line which we found which is the tangent which is seven 𝑦 plus four 𝑥 plus 30 equals zero. And we’ve done this because at the points where they actually meet each other are the points that we’re looking for.

Well, the first stage is actually to rearrange equation two to make 𝑥 the subject. So we’re gonna get four 𝑥 equals negative 30 minus seven 𝑦, so subtract the 30 and seven 𝑦 from each side of the equation. So then, if we divide both sides of our equation by four, we get 𝑥 is equal to negative 30 minus seven 𝑦 over four. So great we’ve now got this and we can actually substitute it back into equation one to help us find what 𝑦 is. So when we do this, we get negative 30 minus seven 𝑦 over four all squared because we’ve substituted that value of negative 30 minus seven 𝑦 over four in for 𝑥 plus 𝑦 squared plus negative 30 minus seven 𝑦 over four plus eight 𝑦 equals zero.

And then, we expand our parentheses. So we get 900 plus 420𝑦 plus 49𝑦 squared over 16 plus 𝑦 squared plus negative 30 minus seven 𝑦 over four plus eight 𝑦 equals zero. So then, what we’re gonna do is multiply through by 16 just to remove the fractions which gives us 900 plus 420𝑦 plus 49𝑦 squared plus 16𝑦 squared minus 120 minus 28𝑦 plus 128𝑦 equals zero. So now, we’re gonna simplify and collect like terms which gives us 65𝑦 squared plus 520𝑦 plus 780 equals zero.

Okay, so we could actually look to factor at this point. But again, can we make it simpler? Always try to simplify and yes, we can. We can actually divide three by 65 which leaves us with a much more manageable 𝑦 squared plus eight 𝑦 plus 12 equals zero. So great, we can now factor this to actually find out our values of 𝑦. So we get 𝑦 plus six multiplied by 𝑦 plus two equals zero. So therefore, 𝑦 is equal to negative six or negative two. So fantastic, we’ve actually now found out our 𝑦-coordinates of the points on the curve that we’re looking for. Now, let’s substitute these values back into one of our equations to find out the value of 𝑥.

So now, what we’re gonna do is actually substitute our 𝑦-values into the equation we made for 𝑥 to find our 𝑥-values. So if we start with 𝑦 equals negative six, we’re gonna get 𝑥 is equal to negative 30 minus seven multiplied by negative six over four which gives us a value of 𝑥 of three. And then, if we substitute in 𝑦 equals negative two, we’re gonna get 𝑥 is equal to negative 30 minus and then seven multiplied by negative two all over four which is gonna give us an 𝑥-value of negative four.

So therefore, we can say that the points on the curve 𝑥 squared plus 𝑦 squared plus 𝑥 plus eight 𝑦 equals zero at which the tangent is perpendicular to the line seven 𝑦 plus four 𝑥 plus 𝑐 equals zero are three, negative six and negative four, negative two.