Video: Finding the Equation of the Tangent to the Curve of a Cubic Function given the Angle the Tangent Makes with the π‘₯-Axis

Find the equation of the tangent to the curve 𝑦 = π‘₯Β³ + 9 π‘₯Β² + 26π‘₯ that makes an angle of 135Β° with the positive π‘₯-axis.

04:14

Video Transcript

Find the equation of the tangent to the curve 𝑦 equals π‘₯ cubed plus nine π‘₯ squared plus 26π‘₯ that makes an angle of 135 degrees with the positive π‘₯-axis.

So we’ve been asked to find the equation of a tangent to a particular curve, which we know we can do using differentiation and the general equation of a straight line. But what does it mean when it says this tangent makes an angle of 135 degrees with the positive π‘₯-axis? Let’s consider a sketch. Well, it will look something like this. The tangent here is shown in pink. And we can see that when intersects with the π‘₯-axis, the angle between the positive π‘₯-axis and the tangent is 135 degrees.

In order to apply the general equation of a straight line 𝑦 minus 𝑦 one equals π‘š π‘₯ minus π‘₯ one, we either need to know the slope π‘š of our line or the coordinates of a point π‘₯ one, 𝑦 one which lies on the line. So how does knowing that our tangent makes an angle of 135 degrees with the positive π‘₯-axis help with determining either of those? Well, the angle on the other side of this line will be 45 degrees because we know that angles on a straight line sum to 180 degrees. We can sketch in a right-angled triangle below this line and recall that the slope of a line is change in 𝑦 over change in π‘₯. That’s the vertical height of this triangle divided by the horizontal distance. But in that right triangle, those sides are the opposite and adjacent in relation to the angle of 45 degrees. So we’re dividing the length of the opposite by the length of the adjacent.

As the line is sloping downwards though, that vertical change is actually the negative of the value of the opposite. So we have that the slope is equal to negative opposite over adjacent. Opposite divided by adjacent defines the tangent ratio. So in fact, this is equal to negative tan of 45 degrees. And tan of 45 degrees is just one. So by considering this right-angled triangle, we found that the slope of this line is negative one. So, we found the slope of our tangent. But we don’t yet know the coordinates of the point on the curve where this tangent is being drawn. To find this, we need to find the point on the curve where the gradient or slope is equal to negative one.

We begin by differentiating the equation of the curve with respect to π‘₯ and applying the power rule of differentiation, giving d𝑦 by dπ‘₯ equals three π‘₯ squared plus 18π‘₯ plus 26. We then set this expression equal to negative one to find the π‘₯-coordinate of the point on the curve, where the gradient is negative one. This simplifies to three π‘₯ squared plus 18π‘₯ plus 27 equals zero. And then dividing through by three gives π‘₯ squared plus six π‘₯ plus nine equals zero. We should notice that this is, in fact, a perfect square. We can write it as π‘₯ plus three all squared. Solving this equation then, this means that π‘₯ plus three must be equal to zero. And so, π‘₯ is equal to negative three.

Next, we need to find the value of 𝑦 when π‘₯ is equal to negative three, which we do by substituting negative three into the equation of the curve. And it gives negative 24. We now know that this tangent has a slope of negative one at the point negative three, negative 24. All that’s left is to substitute into the general equation of the straight line. 𝑦 minus negative 24 equals negative one multiplied by π‘₯ minus negative three. That all simplifies to 𝑦 plus π‘₯ plus 27 equals zero.

The key steps in this question then were to use some trigonometric reasoning to identify that if a line makes an angle of 135 degrees with the positive π‘₯-axis. Then, its gradient or slope is equal to negative tan 45 degrees, which is equal to negative one. We then use the gradient function of the curve to identify the π‘₯-value at which the gradient was equal to negative one. We found the corresponding 𝑦-value by substituting into the equation of the curve and then finally used the general equation of a straight line 𝑦 minus 𝑦 one equals π‘š π‘₯ minus π‘₯ one to find the equation of this tangent.

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