Help with Highschool Calculus Needed

Here is a question I am stuck on.

Find the coordinates of the points on 4x2+y2=124x^2+y^2=12 for which the normal passes through the point (1,2).

#Calculus

Note by Ron Lauterbach
2 years ago

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Comments

Coordinates of point on the ellipse:

(x,α12−4x2)(x, \alpha \sqrt{12 - 4x^2})

Here, α=+1\alpha = +1 corresponds to the positive half of the ellipse, and α=−1\alpha = -1 corresponds to the negative half of the ellipse. If the normal goes through (1,2)(1,2), it follows that the tangent vector to the curve is perpendicular to the vector from (x,y)(x,y) to (1,2)(1,2). Thus, the dot product of those two vectors is zero.

Vector from curve to (1,2)(1,2) (which is also the normal vector):

(1−x,2−α12−4x2)\Big(1-x, 2 - \alpha \sqrt{12 - 4x^2} \Big)

Tangent vector to the curve:

(1,12α112−4x2(−8x))=(1,−4α x12−4x2)\Big(1, \frac{1}{2} \alpha \frac{1}{\sqrt{12 - 4x^2}} (-8x) \Big) \\ = \Big(1, \frac{-4 \alpha \, x}{\sqrt{12 - 4x^2}} \Big)

Setting dot product to zero (recall that α2=1)\alpha^2 = 1):

(1−x,2−α12−4x2)⋅(1,−4α x12−4x2)=01−x−8α x12−4x2+4α2 x=01+3x=8α x12−4x2\Big(1-x, 2 - \alpha \sqrt{12 - 4x^2} \Big) \cdot \Big(1, \frac{-4 \alpha \, x}{\sqrt{12 - 4x^2}} \Big) = 0 \\ 1 - x - \frac{8 \alpha \, x}{\sqrt{12 - 4x^2} } + 4 \alpha^2 \, x = 0 \\ 1 + 3x = \frac{8 \alpha \, x}{\sqrt{12 - 4x^2} } \\

Solving yields:

x≈1.361   for   α=+1x≈−0.188   for   α=−1 x \approx 1.361 \,\,\, \text{for} \,\,\, \alpha = +1 \\ x \approx -0.188 \,\,\, \text{for} \,\,\, \alpha = -1

The graph below plots the normal lines from these points to check that they do in fact go through (1,2) (1,2) :

Steven Chase - 2 years ago

@Ron Lauterbach Does this resolve your issue?

Steven Chase - 2 years ago
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