Question Video: Determining Whether There Is a Common Tangent to Two Given Quadratic Curves Then Finding Its Equation | Nagwa Question Video: Determining Whether There Is a Common Tangent to Two Given Quadratic Curves Then Finding Its Equation | Nagwa

# Question Video: Determining Whether There Is a Common Tangent to Two Given Quadratic Curves Then Finding Its Equation Mathematics • Second Year of Secondary School

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Do the curves π¦ = β2π₯Β² + 4π₯ + 24 and π¦ = β6π₯Β² β 4π₯ + 20 have a common tangent at the point of intersection? If so, give the equation of this tangent.

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### Video Transcript

Do the curves π¦ is equal to negative two π₯ squared plus four π₯ plus 24 and π¦ is equal to negative six π₯ squared minus four π₯ plus 20 have a common tangent at the point of intersection? If so, give the equation of this tangent.

Weβre given two curves defined by quadratic equations. And weβre asked the question, do these curves share a common tangent at the point of intersection? And if they do share a common tangent at the point of intersection, we need to find the equation of this tangent. To answer this question, we should start by finding all of the points of intersections for our two curves. We know for our curves to intersect, their outputs must be equal. So we must have negative two π₯ squared plus four π₯ plus 24 is equal to negative six π₯ squared minus four π₯ plus 20. And we can solve this equation by writing all of our variables on the same side of the equation.

First, we add six π₯ squared to both sides of our equation. We need to subtract two π₯ squared from this, so we get four π₯ squared. Next, we need to add four π₯ to both sides of the equation. And we see we get four π₯ plus four π₯, which is eight π₯. Finally, we subtract 20 from both sides of the equation. We see we have to add 24 giving us four. So we have four π₯ squared plus eight π₯ plus four is equal to zero. We can then divide this equation through by four. This gives us π₯ squared plus two π₯ plus one is equal to zero. This is a quadratic equation. So we can solve this by using a quadratic solver or by using the quadratic formula.

However, we can also just notice this is equal to π₯ plus one all squared. And this will only be equal to zero if π₯ is equal to negative one. So our curves will intersect when π₯ is equal to negative one. We can find the π¦-coordinate of this point by substituting into either about two curves. It doesnβt matter which of our two curves we substitute this in. We get π¦ is equal to negative six times negative one squared minus four times negative one plus 20. And we can just calculate this. We get negative six plus four plus 20 which is equal to 18. So we found the point of intersection of our two curves. Itβs the point negative one, 18.

Now we need to determine if the tangent lines to both of our curves at the point negative one, 18 are the same. To do this, we need to recall the general equation for our line. Itβs given by π¦ minus π¦ one is equal to π times π₯ minus π₯ one, where π is the slope of our straight line and our line passes through the point π₯ one, π¦ one. We want to use this to find the equation of both of our tangent lines. However, we could notice something interesting. Both of our tangent lines pass through the point negative one, 18. So in both cases, π₯ one will be equal to negative one and π¦ one will be equal to 18.

So now the only possible way that our tangent lines can be different is if they have different slopes. So we need to find the slope of each of our two tangent lines. And to do this, we need to remember that the slope of our tangent line at this point will be equal to the slope of our curves at this point. So we need to find the slopes of our curves at the point negative one, 18. And we know to find the slope of a curve at a point, we need to differentiate it with respect to π₯. So we need to differentiate both of our curves with respect to π₯. Letβs start with the first curve, the derivative of negative two π₯ squared plus four π₯ plus 24 with respect to π₯.

We can do this term by term by using the power rule of differentiation. We want to multiply by our exponent of π₯ then reduce this exponent by one. This gives us negative four π₯ plus four. And we can do exactly the same for our second curve. We need to do this term by term by using the power rule for differentiation. We get negative 12π₯ minus four. And we want to find the slope of our curve when π₯ is equal to negative one. So we need to substitute negative one into both of these expressions. Substituting π₯ is equal to negative one into the first derivative, we get π is equal to negative four times negative one plus four, which we can evaluate is equal to eight. And we can do exactly the same for our other derivative. We would get that π is equal to negative 12 times negative one minus four, which we can get a value is also equal to eight.

So in both cases, we can see the slopes of our curve are equal. Therefore, the tangent line to both curves at the point negative one, 18 has the same slopes. They pass through the same point, and they have the same slope. Therefore, they must have the same equation. All thatβs left to do now is substitute π₯ one is equal to negative one, π¦ one is equal to 18, and π is equal to eight into our equation for a straight line. We get π¦ minus 18 is equal to eight times π₯ minus negative one. And we can simplify this. First, subtracting negative one is the same as adding one. Then weβll distribute eight over our parentheses.

This gives us that π¦ minus 18 is equal to eight π₯ plus eight. Finally, weβre going to write all of our terms on the same side of the equation. And by doing this, simplifying, and rearranging, we get our final answer. Yes, they do have a common tangent at the point of intersection. And the equation of this tangent is π¦ minus eight π₯ minus 26 is equal to zero.

Before we end, it can also be useful to see a sketch of these two curves to see exactly whatβs happening. Weβll start by sketching our two parabolas. We can do this in a number of different ways. For example, we could use a graphing calculator, or we could use calculus or completing the square to find the turning point of these two parabolas. We could also use completing the square or a quadratic solver to find the equations of the π₯-intercepts. Finally, we could just substitute π₯ is equal to zero into both of these two equations to find the π¦-intercepts. We would get a sketch, which looks something like this.

And as weβve already seen, thereβs one point of intersection, and we found the coordinates of this point. Itβs the point negative one, 18. And now from our sketch, it looks like these two curves have the same slope at this point. And in fact, we prove that they did and that means that they must have a shared tangent. And we found the equation of this tangent. π¦ minus eight π₯ minus 26 is equal to zero. In this question, we were given two curves defined by quadratic equations. We were able to find the point of intersection of these two quadratics and then prove that they must have a shared tangent at this point of intersection. We were able to show that this shared tangent have the equation π¦ minus eight π₯ minus 26 is equal to zero.

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