# Question Video: Determining Whether Two Curves Defined Implicitly Are Orthogonal or Not Mathematics

Do the curves 9𝑦⁴ − 8𝑦 = 6𝑥 and −5𝑥² − 3𝑦 = −4𝑥 intersect orthogonally at the origin?

05:18

### Video Transcript

Do the curves nine 𝑦 to the fourth power minus eight 𝑦 equals six 𝑥 and negative five 𝑥 squared minus three 𝑦 equals negative four 𝑥 intersect orthogonally at the origin?

Before we start, we can verify that the curves intersect at the origin by checking both equations evaluated at zero, zero. When 𝑥 equals zero and 𝑦 equals zero, we can see that both equations become zero equals zero, which tells us that both curves pass through the origin. Remember, two curves intersect orthogonally at a point if their tangents are well defined at that point and are orthogonal to each other.

Let us find the slopes of the tangents at the origin for each curve by implicitly differentiating the given equations. For the first curve, we have d by d𝑥 of nine 𝑦 to the fourth power minus eight 𝑦 is equal to d by d𝑥 of six 𝑥. The expression on the left-hand side of the equation is in terms of the variable 𝑦. So we need to apply the chain rule. d by d𝑥 of nine 𝑦 to the fourth power minus eight 𝑦 is equal to d by d𝑦 of nine 𝑦 to the fourth power minus eight 𝑦 multiplied by d𝑦 by d𝑥, which equals 36𝑦 cubed minus eight multiplied by d𝑦 by d𝑥.

On the other hand, the right-hand side is d by d𝑥 of six 𝑥, which is equal to six. Hence, we have 36𝑦 cubed minus eight multiplied by d𝑦 by d𝑥 is equal to six. Dividing through by 36𝑦 cubed minus eight, we have d𝑦 by d𝑥 is equal to six over 36𝑦 cubed minus eight. We know that d𝑦 by d𝑥 evaluated at the origin gives the slope of the tangent.

Substituting the origin zero, zero into the equation gives us the following. Calculating the denominator, we have d𝑦 by d𝑥 is equal to six over negative eight, which is equal to negative three-quarters. Thus, the slope of the tangent of the first curve at the origin is negative three-quarters.

Next, let us find the slope of the tangent of the second curve. Implicitly differentiating the second equation gives us d by d𝑥 of negative five 𝑥 squared minus d by d𝑥 of three 𝑦 is equal to d by d𝑥 of negative four 𝑥. The first term on the left-hand side and the term on the right-hand side of the equation are regular derivatives, since the variable of these expressions matches the variable of differentiation 𝑥.

For the second term on the left-hand side, we need to use the chain rule. We have d by d𝑥 of negative five 𝑥 squared is equal to negative 10𝑥. d by d𝑥 of three 𝑦 is equal to d by d𝑦 of three 𝑦 multiplied by d𝑦 by d𝑥, which is equal to three multiplied by d𝑦 by d𝑥. And d by d𝑥 of negative four 𝑥 is equal to negative four. This gives us negative 10𝑥 minus three multiplied by d𝑦 by d𝑥 is equal to negative four.

Rearranging to make d𝑦 by d𝑥 the subject, we have d𝑦 by d𝑥 is equal to negative four plus 10𝑥 divided by negative three. Evaluating this equation at the origin zero, zero and simplifying, we have the following, which simplifies such that d𝑦 by d𝑥 is equal to four-thirds. Thus, the slope of the tangent to the second curve at the origin is four-thirds.

We now have the slopes of both tangents at the origin. Remember, if the slopes of two lines multiply to give negative one, then they are orthogonal. We obtained the slopes of the tangents for these two curves: negative three-quarters and four-thirds. And we see that negative three-quarters multiplied by four-thirds is equal to negative one.

We can therefore conclude that the two curves intersect orthogonally at the origin.