At a point on the curve 𝑥 squared plus three 𝑥 plus 𝑦 squared plus five 𝑦 plus four equals zero with 𝑥 is less than zero, 𝑦 is less than zero, the tangent makes an angle of nine 𝜋 by four with the positive 𝑥-axis. Find the equation of the tangent at that point.
We’ve been given an equation of a curve in terms of 𝑥 and 𝑦. We’re also told some information about the angle that the tangent makes with the positive 𝑥-axis. Now, before we do anything, let’s actually think what it means for the tangent to make an angle of nine 𝜋 by four radians with the positive 𝑥-axis. The positive 𝑥-axis is here. A full turn, 360 degrees, is a turn of two 𝜋 radians or eight 𝜋 by four radians. Since the tangent makes an angle of nine 𝜋 by four radians with the positive 𝑥-axis, we need to travel another 𝜋 by four radians from that full turn.
Remember, 𝜋 by four radians is equivalent to 45 degrees. And so, we can construct a right-angled triangle on our tangent. We know that this triangle must be isosceles since two of its angles are 45 degrees. This means if we were to calculate rise over run or change in 𝑦 divided by change in 𝑥, we’d end up finding that the slope is equal to one. And so, given the information about how a tangent, we know its slope, but we need to find a point that it passes through. So how are we going to do that?
Well, we’re going to use the equation of the curve. We know that the slope of the tangent is found by evaluating the derivative at a given point. And we’ve been given an equation that’s defined implicitly. So, we’re going to use implicit differentiation to differentiate both sides of our equation with respect to 𝑥. Remember, this is just a special case of the chain rule. And it says that to differentiate a function in 𝑦 with respect to 𝑥, we differentiate that function with respect to 𝑦, and then multiply that by d𝑦 by d𝑥.
Let’s differentiate term by term. We begin by differentiating 𝑥 squared with respect to 𝑥. Well, remember to differentiate a power term, we multiply the entire term by the exponent and reduce that exponent by one. So, the derivative of 𝑥 squared with respect to 𝑥 is two 𝑥. Next, we’ll differentiate three 𝑥 with respect to 𝑥, and we get three. But what about the third term, the derivative of 𝑦 squared with respect to 𝑥? Well, remember, we begin by differentiating that with respect to 𝑦. And the derivative of 𝑦 squared with respect to 𝑦 is two 𝑦. We then multiply that by d𝑦 by d𝑥. So, our third term is two 𝑦 d𝑦 by d𝑥.
We repeat this process for the derivative of five 𝑦 with respect to 𝑥. We differentiate it with respect to 𝑦, that’s five, and then times that by d𝑦 by d𝑥. Finally, we know that the derivative of any constant is zero. And so, we find two 𝑥 plus three plus two 𝑦 d𝑦 by d𝑥 plus five d𝑦 by d𝑥 must be equal to zero. Our next job is to factor d𝑦 by d𝑥. When we do, we find that our equation becomes d𝑦 by d𝑥 times two 𝑦 plus five plus two 𝑥 plus three equals zero.
Now, we want to express d𝑦 by d𝑥 as some function in 𝑥 and 𝑦. So, we subtract two 𝑥 and three from both sides and then divide through by two 𝑦 plus five. So, we’ve got an expression for the derivative of 𝑦 with respect to 𝑥. It’s negative two 𝑥 minus three ever two 𝑦 plus five.
Now, remember, we said that the slope of our tangent was one, so let’s set this equal to one. Our aim is to find an expression for 𝑦 in terms of 𝑥. So, we multiply through by two 𝑦 plus five. Subtract five from both sides to get two 𝑦 equals negative two 𝑥 minus eight and then divide through by two. So, we get 𝑦 equals negative 𝑥 minus four. But why did we do this? Well, we can now substitute 𝑦 equals negative 𝑥 minus four into our original equation. We get 𝑥 squared plus three 𝑥 plus negative 𝑥 minus four all squared plus five times negative 𝑥 minus four plus four equals zero.
Distributing the parentheses and we see that negative 𝑥 minus four squared is the same as 𝑥 squared plus eight 𝑥 plus 16, and five times negative 𝑥 minus four is negative five 𝑥 minus 20. This whole expression on the left-hand side simplifies to two 𝑥 squared plus six 𝑥. And we can factor the expression on the left to help us solve for 𝑥. When we do, we get two 𝑥 times 𝑥 plus three equals zero.
Well, we know that one of the solutions to this equation is found by solving two 𝑥 equals zero. So, 𝑥 equals zero. The other is the value of 𝑥 such that 𝑥 plus three is equal to zero. Well, that’s 𝑥 equals negative three. But of course, we were told that 𝑥 is less than zero. So, we’re going to choose 𝑥 equals negative three. We use the equation 𝑦 equals negative 𝑥 minus four to find the corresponding value of 𝑦. And we get 𝑦 equals negative negative three minus four, which is equal to negative one. And we now know the gradient or the slope of our tangent is one, and we know it passes through the point with coordinates negative three, negative one.
Our last job is to substitute everything we know into the equation of a straight line. We get 𝑦 minus 𝑦 one equals 𝑚 times 𝑥 minus 𝑥 one. We get 𝑦 minus negative one equals one times 𝑥 minus negative three. Distributing our parentheses and we get 𝑦 plus one equals 𝑥 plus three. And then we subtract 𝑥 and three from both sides.
And the equation of the tangent to our curve at the point we calculated to be negative three, negative one is, therefore, negative 𝑥 plus 𝑦 minus two equals zero.