Question Video: Finding the Equation of the Normal to a Curve Defined Implicitly at a Given Point Mathematics

Determine the equation of the line normal to the curve 4𝑦² = 𝑥⁴ at the point (2, −2).

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Video Transcript

Determine the equation of the line normal to the curve four 𝑦 squared is equal to 𝑥 to the fourth power at the point two, negative two.

In this question, we’re asked to determine the equation of the line which is normal to the curve four 𝑦 squared is equal to 𝑥 to the fourth power at the point two, negative two. Since the question wants us to find the equation of a line, let’s start by recalling the general equation for a line. This is given as 𝑦 minus 𝑦 one is equal to 𝑚 multiplied by 𝑥 minus 𝑥 one, where 𝑚 is the slope of our line and our line passes through the point 𝑥 one, 𝑦 one. And, in this question, we’re actually given a point that our line passes through. We’re told that it passes through the point two, negative two. So we can set our value of 𝑥 one equal to two and our value of 𝑦 one equal to negative two.

This means all we need to do to find the equation of our normal line is to find the value of our slope 𝑚. And this is where we run into a problem. We can see we’re not given 𝑦 as some function in 𝑥; instead, we’re given four 𝑦 squared is equal to 𝑥 to the fourth power. So we can’t just directly find an expression for d𝑦 by d𝑥. We’re going to need to do something different. There’s several different options we could do here. One way would be to rearrange the equation for our curve to give 𝑦 as a function of 𝑥. We can start by dividing both sides of our equation through by four. This gives us that 𝑦 squared is equal to 𝑥 to the fourth power over four. We want to rearrange this equation to give 𝑦 as a function in 𝑥, so we’re going to want to take the square root of both sides of this equation.

However, we do need to be careful here. Remember, we’re going to get a positive and a negative square root. This gives us that 𝑦 is equal to positive or negative the square root of 𝑥 to the fourth power over four. And at this point, we might be worried because we don’t know how to differentiate an expression with a positive and a negative sign. However, we’re trying to find the equation of our normal line to our curve at the point two, negative two. And to do this, all we need to know is the slope of our curve at the point two, negative two. And at this point, we can see the 𝑦-coordinate is equal to negative two. This means we have to take the negative square root. So at this point, we must have that 𝑦 is equal to negative the square root of 𝑥 to the fourth power over four.

And of course, we can simplify this by evaluating the square root of the numerator and our denominator. This gives us that 𝑦 is equal to negative 𝑥 squared over two. And it’s worth pointing out that this will only be true if our values of 𝑦 are less than or equal to zero, which is true for the point we’re interested in. Now, we’re going to need to recall how do we find the slope of a normal line to a curve at a point. Well, the normal line to the curve at this point must be perpendicular to the tangent line to the curve at this point. So instead of directly finding the slope of our normal line, let’s instead find the slope of the tangent line at this point.

And remember, to find the slope of a tangent line to a curve at a point, we need to find an expression for d𝑦 by d𝑥. So we need to differentiate our curve with respect to 𝑥. We get d𝑦 by d𝑥 is equal to the derivative of negative 𝑥 squared over two with respect to 𝑥. And we can evaluate the derivative of this expression by using the power rule for differentiation. We want to multiply by our exponent of 𝑥 and reduce this exponent by one. This gives us negative two 𝑥 to the first power over two. And we can simplify this to just give us negative 𝑥.

Now, we want to substitute the value of 𝑥 is equal to two into this expression. This gives us the slope of the tangent line to the curve four 𝑦 squared is equal to 𝑥 to the fourth power at the point two, negative two is negative two. But remember, we’re not looking for the slope of the tangent line at this point. We want the slope of the normal line at this point. And to do this, all we need to recall is that to find the slope of a normal line to a curve at a point given the slope of its tangent line, we just take negative the reciprocal of this value. In other words, the slope of our normal line is going to be negative one divided by negative two. And we can evaluate this; it’s just equal to one-half.

And there’s a few things worth pointing out here. First, if the slope of our tangent line had been equal to zero, we couldn’t have used this formula to find the slope of our normal line. In this case, if the slope of our tangent line was zero, this would tell us that our tangent line is horizontal. And if our tangent line is horizontal, then, for our normal line to be perpendicular to this, it must be vertical. We could then use this to find the equation of our normal line. In fact, we could say something very similar in the opposite direction. If our tangent line was vertical, then we would also know that our normal line would have to be horizontal. But this isn’t the case we have in this question.

All we need to do now is substitute our values of 𝑥 one, 𝑦 one and the slope 𝑚 into the equation for our line. Substituting in 𝑥 one is equal to two, 𝑦 one is equal to negative two, and 𝑚 is equal to one-half into the equation for our straight line, we get 𝑦 minus negative two is equal to one-half multiplied by 𝑥 minus two is the equation of our normal line. And we can simplify this expression. First, on the left-hand side of our equation, subtracting negative two is the same as adding two. Next, on the right-hand side of our equation, we’re going to distribute one-half over our parentheses. This gives us 𝑥 over two minus one. The last thing we’re going to do is subtract two from both sides of our equation. And this gives us our final answer of 𝑦 is equal to 𝑥 over two minus three.

Therefore, we were able to find the equation of the line normal to the curve four 𝑦 squared is equal 𝑥 to the fourth power at the point two, negative two. We were able to show its equation is 𝑦 is equal to 𝑥 over two minus three.

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