# Question Video: Finding the Value of the Parameter That Makes a Curve Defined by Parametric Equations Have a Horizontal Tangent Mathematics • Higher Education

A curve has parametric equations π₯ = 7πΒ³ + 5πΒ² + π + 4 and π¦ = 6πΒ² β 6π β 8. Find π for which the tangent is horizontal.

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### Video Transcript

A curve has parametric equations π₯ is equal to seven π cubed plus five π squared plus π plus four and π¦ is equal to six π squared minus six π minus eight. Find π for which the tangent is horizontal.

In this question, we are given parametric equations where our coordinates π₯ and π¦ are given in terms of π. We know that when the tangent is horizontal, its gradient or slope, dπ¦ by dπ₯, must be equal to zero. We also know that dπ¦ by dπ₯ is equal to dπ¦ by dπ divided by dπ₯ by dπ. This can also be written using the chain rule as dπ¦ by dπ multiplied by dπ by dπ₯. We multiply dπ¦ by dπ by the reciprocal of dπ₯ by dπ.

Letβs begin by considering π₯, which is equal to seven π cubed plus five π squared plus π plus four. We can differentiate this term by term to get an expression for dπ₯ by dπ. Differentiating seven π cubed gives us 21π squared, differentiating five π squared gives us 10π, and differentiating π gives us one. When we differentiate the constant four, we get zero. Therefore, dπ₯ by dπ is equal to 21π squared plus 10π plus one.

We can repeat this process for π¦, which is equal to six π squared minus six π minus eight. dπ¦ by dπ is equal to 12π minus six. We can then find an expression for dπ¦ by dπ₯ by dividing dπ¦ by dπ by dπ₯ by dπ. dπ¦ by dπ₯ is equal to 12π minus six divided by 21π squared plus 10π plus one.

For the tangent to be horizontal, we know that dπ¦ by dπ₯ must equal zero. This means that we need to solve this equation equal to zero. As the denominator cannot be equal to zero, the numerator must be. 12π minus six must equal zero. Adding six to both sides gives us 12π is equal to six. Dividing both sides of this equation by 12 gives us π is equal to six over 12 or six twelfths. This fraction is equivalent to one-half. The value of π for which the tangent is horizontal is one-half.

Whilst we donβt need to in this question, we could then substitute this value for π back into our equations for π₯ and π¦. This would give us the coordinate of the point at which the tangent is horizontal.