Video Transcript
The curves 𝑦 equals two 𝑥 squared minus three 𝑥 minus two and 𝑦 equals negative three 𝑥 squared plus five 𝑥 minus five intersect orthogonally at a point. What is this point?
We’ll begin by recalling what it means for two curves to intersect orthogonally. Firstly, the two curves need to meet or cross at this point, but secondly the tangents to the two curves must be perpendicular to one another. Suppose the two tangents at this point have the equations 𝑦 equals 𝑚 one 𝑥 plus 𝑏 one and 𝑦 equals 𝑚 two 𝑥 plus 𝑏 two. Well, we know that if two lines are perpendicular, then the product of their slopes is negative one. So 𝑚 one multiplied by 𝑚 two must be negative one. We also know that the slope of a tangent to a curve at any given point is the same as the slope of the curve itself at that point. And we can find the slope function of a curve using differentiation.
We’ll begin then by finding the slope function for each of these curves. Now, each curve is a polynomial. In fact, they are each quadratics in 𝑥. And so we can differentiate each equation by recalling the general power rule of differentiation. This tells us that for real values of the constants 𝑎 and 𝑛, the derivative with respect to 𝑥 of 𝑎 multiplied by 𝑥 to the 𝑛th power is 𝑎 multiplied by 𝑛 multiplied by 𝑥 to the 𝑛 minus first power. So the slope function for the first curve is d𝑦 by d𝑥 equals four 𝑥 minus three and for the second d𝑦 by d𝑥 is equal to negative six 𝑥 plus five. And in each case, we recall that the derivative of a constant with respect to 𝑥 is simply zero.
Now, in order to find the coordinates of the point at which these two curves intersect orthogonally, we first need to find any 𝑥-values at which the product of their slopes is negative one. We’re looking to solve the equation four 𝑥 minus three multiplied by negative six 𝑥 plus five is equal to negative one. Distributing the parentheses, we have negative 24𝑥 squared plus 20𝑥 plus 18𝑥 minus 15 is equal to negative one. We can group the like terms, so we can combine positive 20𝑥 and positive 18𝑥 on the left-hand side to make positive 38𝑥. And then, we can add one to each side to make a constant of negative 14 on the left-hand side and zero on the right-hand side. So our equation becomes negative 24𝑥 squared plus 38𝑥 minus 14 is equal to zero.
We can simplify this quadratic by dividing through by negative one as this will give a positive coefficient of 𝑥 squared. And as all of the coefficients are multiples of two, we can also divide through by two. So we change the sign of each term and also halve its coefficient. So we now have 12𝑥 squared minus 19𝑥 plus seven is equal to zero.
Let’s see if we can solve this equation by factoring. As the constant term in the equation is positive seven, which is a prime number, we know that we need the numbers seven and one, which are the only factors of seven in our parentheses. The coefficient of the 𝑥-term, though, is negative, which means we need the numbers to in fact be negative seven and negative one, which still have a product of positive seven.
The first term in each set of parentheses need to be a pair of 𝑥-terms that will multiply to give 12𝑥 squared. Now, these could be 12𝑥 and 𝑥, six 𝑥 and two 𝑥, or four 𝑥 and three 𝑥. And we can find the correct pair by using trial and error. When we do, we find that we need 12𝑥 in the first set of parentheses with the negative seven and 𝑥 or one 𝑥 in the second set of parentheses with the negative one, so the quadratic factors as 12𝑥 minus seven multiplied by 𝑥 minus one. We can check that we do get the correct coefficient of 𝑥 when we redistribute the parentheses. Negative seven multiplied by 𝑥 gives negative seven 𝑥 and 12𝑥 multiplied by negative one gives negative 12𝑥. So we have negative seven 𝑥 and negative 12𝑥, which does make negative 19𝑥.
Next, we recall that if a product is equal to zero, then one of the individual factors must be equal to zero. So we have either 12𝑥 minus seven equals zero or 𝑥 minus one equals zero. The solution to the first equation is 𝑥 equals seven over 12. And the solution to the second equation is 𝑥 equals one. So we have two 𝑥-values at which the slopes of these two curves multiply to give negative one. But this doesn’t automatically mean that the curves intersect orthogonally at these points. We must also check that the curves also intersect each other at each of these 𝑥-values, which means we need to check that the 𝑦-value for each curve is the same for each of these given 𝑥-values.
When 𝑥 is equal to one, the 𝑦-value for the first curve will be two multiplied by one squared minus three multiplied by one minus two, which is equal to negative three. For the second curve, the 𝑦-value is negative three multiplied by one squared plus five multiplied by one minus five, which is also equal to negative three. As the 𝑦-values are the same when 𝑥 equals one, this confirms that the two curves do intersect at this value of 𝑥. Let’s check the other value of 𝑥 though. When 𝑥 is equal to seven over 12, the 𝑦-value for the first curve is two multiplied by seven over 12 squared minus three multiplied by seven over 12 minus two. That’s negative 221 over 72 or as a decimal negative 3.06 continuing.
For the second curve, we have 𝑦 equals negative three multiplied by seven over 12 squared plus five multiplied by seven over 12 minus five. That’s negative 149 over 48 or as a decimal negative 3.10 continuing. As the 𝑦-values are not the same when 𝑥 equals seven over 12, the two curves do not intersect at this value. So we found that the coordinates of the point at which these two curves not only intersect but they also intersect orthogonally so their tangents are perpendicular is the point with coordinates one, negative three.
