Video: Finding the Equations of the Normal to the Curve of a Quadratic Function at the Points of Intersection of the Curve with Another Given Line

List the equations of the normals to 𝑦 = π‘₯Β² + 2π‘₯ at the points where the curve meets the line 𝑦 βˆ’ 4π‘₯ = 0.

03:05

Video Transcript

List the equations of the normals to 𝑦 equals π‘₯ squared plus two π‘₯ at the points where the curve meets the line 𝑦 minus four π‘₯ equals zero.

What is meant by the term normal in this context? Well, we recall first of all that the tangent to a curve has the same gradient as the curve at that point. The normal, however, passes through that same point, but it is perpendicular to the tangent at that point. We can use properties of perpendicular lines to deduce the relationship that exists between the gradient of the tangent and the gradient of the normal to a curve at a given point. The product of the two gradients will be equal to negative one and they will be negative reciprocals of one another.

We must make sure that we’re clear whether we’ve been asked to find the equation of a tangent or a normal when we’re answering questions like this. So now that we know what normals are, let’s see how we can answer this question. We’ve been asked to list the equations of the normals to a given curve at the point where this curve meets another line. So our first step is going to be to find these points of intersection.

We can rearrange the equation of the line to give 𝑦 equals four π‘₯ and then set the two expressions for 𝑦 equal to one another to give an equation in π‘₯ only. We can subtract four π‘₯ from each side and then factor the resulting quadratic to give π‘₯ multiplied by π‘₯ minus two is equal to zero. The two roots of this equation are π‘₯ equals zero or π‘₯ equals two. So we know the π‘₯-coordinates of our points of intersection. To find the corresponding 𝑦-coordinates, we substitute each π‘₯-value back into the equation of the curve to give 𝑦 equals zero when π‘₯ equals zero and 𝑦 equals eight when π‘₯ equals two.

So we now know the two points of intersection. And we, therefore, know the coordinates of one point that lies on each normal. But we need to determine the gradient or slope of each normal. First, we can find the slope of each tangent by differentiating 𝑦 with respect to π‘₯, giving d𝑦 by dπ‘₯ equals two π‘₯ plus two. When π‘₯ equals zero, the slope will be two. And when π‘₯ equals two, the slope will be six. But remember, this is the slope of the tangent, not the slope of the normal. To find the slope of each normal, we need to take the negative reciprocal of the slope of each tangent. So the slope of our first normal is negative a half and the slope of our second is negative one-sixth.

Finally, we can apply the formula for the general equation of a straight line. For the first normal with a slope of negative a half passing through the point zero, zero, we get the equation two 𝑦 plus π‘₯ equals zero. And for the second with a slope of negative one-sixth passing through the point two, eight, we get the equation six 𝑦 plus π‘₯ minus 50 equals zero. So we found the equations of the two normals. We must be really careful on questions like this. Remember, the slope of the normal is not the same as the slope of the tangent. It’s equal to the negative reciprocal of the slope of the tangent because the two lines are perpendicular to one another.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.