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

At the point (0, βˆ’2), determine the equation of the normal to the curve represented by the equation 6π‘₯Β³ + 2π‘₯Β² + 2π‘₯ βˆ’ 9𝑦² βˆ’ 8𝑦 + 20 = 0.

06:02

Video Transcript

At the point zero, negative two, determine the equation of the normal to the curve represented by the equation six π‘₯ cubed plus two π‘₯ squared plus two π‘₯ minus nine 𝑦 squared minus eight 𝑦 plus 20 is equal to zero.

The question wants us to determine the equation of the normal to the curve defined by the implicit equation six π‘₯ cubed plus two π‘₯ squared plus two π‘₯ minus nine 𝑦 squared minus eight 𝑦 plus 20 is equal to zero at the point zero, negative two. To answer this question, we first need to remember what it means for a line to be normal to the curve.

We recall we call a line normal to the curve at a point if it’s perpendicular to the tangent to the curve at that point. And we know if the tangent to the curve at that point has a slope of π‘š, then the normal to the curve at that point must have a slope of negative one divided by π‘š. It’s also worth noting at this point if our tangent line was horizontal, then our normal line will be vertical. So, technically, its slope would be undefined. And the same is true in reverse. If our normal line was horizontal, then our tangent line would be vertical.

So to find the equation of our normal to the curve at the point zero, negative two, we need to find the slope of the tangent to the curve at the point zero, negative two. Since we know d𝑦 by dπ‘₯ tells us the slope of the tangent line, we can find the slope of the tangent line by differentiating both sides of our implicit equation with respect to π‘₯. This gives us the derivative of six π‘₯ cubed plus two π‘₯ squared plus two π‘₯ minus nine 𝑦 squared minus eight 𝑦 plus 20 with respect to π‘₯ is equal to the derivative of zero with respect to π‘₯.

We can evaluate this by differentiating each term individually. We know the derivative with respect to π‘₯ of zero is just equal to zero. We can find the derivative of six π‘₯ cubed, two π‘₯ squared, two π‘₯, and 20 with respect to π‘₯ by using the power rule for differentiation. We multiply by the exponent and then reduce the exponent by one. To differentiate negative nine 𝑦 squared and negative eight 𝑦, we need to notice that 𝑦 is a function of π‘₯.

Since 𝑦 is a function of π‘₯, we can differentiate this using the chain rule. This gives us the derivative of 𝑓 with respect to π‘₯ is equal to the derivative of 𝑓 with respect to 𝑦 multiplied by the derivative of 𝑦 with respect to π‘₯. This gives us the derivative of negative nine 𝑦 squared with respect to π‘₯ is equal to the derivative of negative nine 𝑦 squared with respect to 𝑦 multiplied by d𝑦 by dπ‘₯. And we can differentiate negative nine 𝑦 squared with respect to 𝑦 by using the power rule for differentiation. This gives us negative 18𝑦.

We can do the same to differentiate negative eight 𝑦 with respect to π‘₯. This gives us the derivative of negative eight 𝑦 with respect to 𝑦 times d𝑦 by dπ‘₯. And we can differentiate negative eight 𝑦 with respect to 𝑦 by using the power rule for differentiation. It’s equal to negative eight. So, we now have 18π‘₯ squared plus four π‘₯ plus two minus 18𝑦 d𝑦 by dπ‘₯ minus eight d𝑦 by dπ‘₯ plus zero is equal to zero.

And we want to find the slope of our tangent line at the point zero, negative two so that we can find the slope of the normal at this point. To do this, we need to rearrange this equation so that d𝑦 by dπ‘₯ is the subject. We’ll subtract 18π‘₯, four π‘₯, and two from both sides of this equation. And we’ll factor out d𝑦 by dπ‘₯ from the remaining terms on the left-hand side of our equation. This gives us negative 18𝑦 minus eight multiplied by d𝑦 by dπ‘₯ is equal to negative 18π‘₯ squared minus four π‘₯ minus two.

We can then make d𝑦 by dπ‘₯ the subject of this equation by dividing both sides of our equation by negative 18𝑦 minus eight. This gives us d𝑦 by dπ‘₯ is equal to negative 18π‘₯ squared minus four π‘₯ minus two all divided by negative 18𝑦 minus eight. We can simplify this by canceling the shared factor of negative two in our numerator and our denominator, giving us nine π‘₯ squared plus two π‘₯ plus one all divided by nine 𝑦 plus four.

Since d𝑦 by dπ‘₯ tells us the slope of our tangent, we can find the slope of the tangent to the curve at the point zero, negative two by substituting in π‘₯ is equal to zero and 𝑦 is equal to negative two. This gives us the slope of our tangent to the curve at the point zero, negative two is equal to nine times zero squared plus two times zero plus one all divided by nine times negative two plus four, which we can calculate to give us negative one divided by 14.

However, this is the slope of our tangent. We can find the slope of the normal by taking negative one times the reciprocal of this value. So, we’ve shown the slope of our normal to the curve at the point zero, negative two is negative one times the reciprocal of negative one divided by 14, which is just equal to 14. However, the question wants us to find the equation of the normal to the curve at this point. There are several different ways of finding the equation for this line. For example, we can write our line in the form 𝑦 is equal to the slope of the line times π‘₯ plus the 𝑦-intercept, which we will call 𝑐.

In fact, we’re finding the normal to the curve at the point zero, negative two. This means it passes through the point zero, negative two. And if our line passes through the point zero, negative two which lies on the 𝑦-axis, that means its 𝑦-intercept is negative two. So, 𝑐 is equal to negative two. And the equation to the normal of the curve at the point zero, negative two is 𝑦 is equal to 14π‘₯ minus two. We can rearrange this to get 𝑦 minus 14π‘₯ plus two is equal to zero.

Therefore, we’ve shown the equation of the normal line to the curve represented by the equation six π‘₯ cubed plus two π‘₯ squared plus two π‘₯ minus nine 𝑦 squared minus eight 𝑦 plus 20 is equal to zero is given by 𝑦 minus 14π‘₯ plus two is equal to zero.

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