Parametric Equations

The parametric equations of a curve are:

x = cos\theta, y= 2sin\theta

Find the equation of the normal to the curve at the point

P(cos\theta,2sin\theta)

\displaystyle \frac{dy}{dx} = \frac{2cos\theta}{-sin\theta}

gdt =\displaystyle \frac{-sin\theta}{2cos\theta}

y - 2sin\theta = \displaystyle \frac{-sin\theta}{2cos\theta} (x - cos\theta)

\displaystyle \frac{2y}{sin\theta} - 4 = \frac{-1}{cos\theta}(x - cos\theta)

\displaystyle \frac{2y}{sin\theta} + \frac{x}{cos\theta} = 5

But the book says:

\displaystyle \frac{2y}{sin\theta} - \frac{x}{cos\theta} = 3

Have I went wrong anywhere or is the answer in the book incorrect?

when looking for the perpendicular gradient you need the negative reciprocal

gdt =\displaystyle \frac{-sin\theta}{2cos\theta}

Gradient of a Normal is - dx/dy, you need to remove a factor of -1 from the right hand side

when looking for the perpendicular gradient you need the negative reciprocal

No valid excuse for forgetting this. Funnily enough I done it right in the question before. Thank you.

No valid excuse for forgetting this. Funnily enough I done it right in the question before. Thank you.

Easily done :smile:

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