# Question Video: Finding Coordinates of Points on a Curve, and Using Implicit Differentiation to Find the Equation of a Tangent Mathematics

The equation π¦ Β² β 24π₯Β³ + 24π₯ = 0 describes a curve in the plane. Find the coordinates of two points on this curve where π₯ = β1/2. Determine the equation of the tangent at the point where π₯ = β1/2 and the π¦-coordinate is positive. Find the coordinates of another point, if it exists, at which the tangent meets the curve.

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### Video Transcript

The equation π¦ squared minus 24π₯ cubed plus 24π₯ equals zero describes a curve in the plane. Find the coordinates of two points on this curve where π₯ is equal to negative one-half. Determine the equation of the tangent at the point where π₯ equals negative one-half and the π¦-coordinate is positive. Find the coordinates of another point, if it exists, at which the tangent meets the curve.

There are three parts to this question, the first of which asks us to identify two points that lie on the curve where π₯ is equal to negative one-half. We are told that the equation of the curve is π¦ squared minus 24π₯ cubed plus 24π₯ equals zero. This means that in order to find coordinates where π₯ equals negative one-half, we can substitute this into the equation.

When π₯ equals negative one-half, we have π¦ squared minus 24 multiplied by negative a half cubed plus 24 multiplied by negative a half is equal to zero. Negative one-half cubed is equal to negative one-eighth. And multiplying this by negative 24 gives us positive three. Since 24 multiplied by negative one-half is negative 12, the equation simplifies to π¦ squared plus three minus 12 equals zero. This in turn simplifies to π¦ squared minus nine equals zero. We can then add nine to both sides. And taking the square root of both sides of this equation, we have π¦ is equal to positive or negative root nine, which means that π¦ is equal to positive or negative three.

These are the two π¦-coordinates that have a corresponding π₯-coordinate equal to negative one-half. And so, the two coordinates required are negative one-half, three and negative one-half, negative three. We will now clear some space and answer the second part of this question.

This part wants us to determine the equation of the tangent at the point where π₯ equals negative one-half and the π¦-coordinate is positive. From the first part to this question, we see that this occurs when π₯ equals negative one-half and π¦ is equal to three. So, how do we find the equation of a tangent? Well, the tangent is simply a straight line. So, if we can calculate its slope or gradient π and we know a point through which it passes, π₯ sub one, π¦ sub one, then we can use the general equation of a straight line. This states that π¦ minus π¦ sub one is equal to π multiplied by π₯ minus π₯ sub one.

As the point passes through the point with coordinates negative one-half, three, these will be our values of π₯ sub one and π¦ sub one. We therefore just need to calculate the slope of the tangent to the curve. This slope is found by calculating the derivative dπ¦ by dπ₯ and evaluating that at π₯ is equal to negative one-half.

Now, we have an equation π¦ squared minus 24π₯ cubed plus 24π₯ is equal to zero, and weβre therefore going to need to differentiate each part of this equation with respect to π₯. Now, the first part of this equation is π¦ squared. So, weβre going to use implicit differentiation to differentiate this with respect to π₯. This is a special version of the chain rule. To differentiate π¦ squared, we differentiate it with respect to π¦. That gives us two π¦. And then, we multiply this by dπ¦ by dπ₯. So, the derivative of π¦ squared with respect to π₯ is two π¦ multiplied by dπ¦ by dπ₯.

Next, weβll differentiate negative 24π₯ cubed. We multiply by the exponent and then reduce that exponent by one, giving us negative 72π₯ squared. The derivative of 24π₯ is 24. And differentiating the constant zero, we simply get zero. The equation for the derivative is therefore two π¦ multiplied by dπ¦ by dπ₯ minus 72π₯ squared plus 24 equals zero. From here, we need to make dπ¦ by dπ₯ the subject. In order to do this, we add 72π₯ squared and subtract 24 from both sides. We can then divide through by two π¦ such that dπ¦ by dπ₯ is equal to 72π₯ squared minus 24 all over two π¦.

We are now in a position to substitute our values of π₯ and π¦. We have π₯ is equal to negative one-half and π¦ is equal to three. The slope of the tangent at the given point is therefore equal to 72 multiplied by negative one-half squared minus 24 all divided by two multiplied by three. This simplifies to negative six over six, which is equal to negative one. We now have the values of π together with π₯ sub one and π¦ sub one as required.

The equation of the tangent is therefore equal to π¦ minus three is equal to negative one multiplied by π₯ minus negative one-half. Distributing the parentheses on the right-hand side, we get negative π₯ minus one-half. We can then add three to both sides, giving us a final equation of π¦ is equal to five over two minus π₯. We have therefore found the equation of the tangent as required. At the point negative one-half, three, the equation of the tangent is π¦ is equal to five over two minus π₯.

We will now move on to the third and final part of this question. This part asks us to identify any further coordinates of points at which the tangent meets the curve. We already know that this tangent meets the curve at the point negative one-half, three. So, how do we find out if any other points exist? Well, we can take the equation π¦ is equal to five over two minus π₯ and substitute this into the equation of the curve. Specifically, weβre going to replace π¦ with five over two minus π₯. When we do this, we get five over two minus π₯ all squared minus 24π₯ cubed plus 24π₯ is equal to zero.

Distributing the parentheses gives us 25 over four minus five π₯ plus π₯ squared. And the equation simplifies as shown. We can then collect like terms such that negative 24π₯ cubed plus π₯ squared plus 19π₯ plus 25 over four is equal to zero. So, how do we solve this cubic equation? We could, of course, use a solver on a calculator. But we also know one of the factors already.

By the factor theorem, since negative one-half is a solution to this equation, then π₯ plus a half must be a factor. We could then use polynomial long division to calculate the other factor. When we do this, we find that we can write this equation as π₯ plus a half multiplied by negative 24π₯ squared plus 13π₯ plus 25 over two is equal to zero. We can then use any method for solving quadratics to solve the equation negative 24π₯ squared plus 13π₯ plus 25 over two is equal to zero.

There are two solutions to this equation: π₯ equals 25 over 24 and π₯ is equal to negative one-half. We have already established that π₯ equals negative one-half is a solution. In fact, it appears to be a repeated root. We therefore need to substitute π₯ equals 25 over 24 into our equation for the tangent. Clearing some space, we have π¦ is equal to five over two minus 25 over 24. And this is equal to 35 over 24. So, the coordinates of the other point at which the tangent meets the curve is 25 over 24, 35 over 24.

And we now have answers to all three parts of the question. The coordinates of the two points on the curve where π₯ is equal to negative one-half are negative one-half, three and negative one-half, negative three. The equation of the tangent at the point where π₯ equals negative one-half and the π¦-coordinate is positive is π¦ is equal to five over two minus π₯. And the coordinates of the other point at which the tangent meets the curve is 25 over 24, 35 over 24.

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