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ziro4ka [17]
3 years ago
15

At which points are the tangents drawn to the ellipse x 2 + y 2 = [ a ] x + [ a ] y parallel to

Mathematics
1 answer:
In-s [12.5K]3 years ago
3 0

The given equation of the ellipse is x^2 + y^2 = 2 x + 2 y

At tangent line, the point is horizontal with the x-axis therefore slope = dy / dx = 0

<span>So we have to take the 1st derivative of the equation then equate dy / dx to zero.</span>

x^2 + y^2 = 2 x + 2 y

x^2 – 2 x = 2 y – y^2

(2x – 2) dx = (2 – 2y) dy

(2x – 2) / (2 – 2y) = 0

2x – 2 = 0

x = 1

 

To find for y, we go back to the original equation then substitute the value of x.

x^2 + y^2 = 2 x + 2 y

1^2 + y^2 = 2 * 1 + 2 y

y^2 – 2y + 1 – 2 = 0

y^2 – 2y – 1 = 0

Finding the roots using the quadratic formula:

y = [-(- 2) ± sqrt ( (-2)^2 – 4*1*-1)] / 2*1

y = 1 ± 2.828

y = -1.828 , 3.828

 

<span>Therefore the tangents are parallel to the x-axis at points (1, -1.828) and (1, 3.828).</span>

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<h2>Answer:</h2>\frac{11y-110}{50-5y}=-\frac{11}{5}

<h2>Explanations:</h2>

Given the following expression show below;

\frac{11y-110}{50-5y}

Factor out the GCF from both the numerator and denominator

\frac{11(y-10)}{5(10-y)}

This can also be expressed as;

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Cancel out the common expression

\frac{-11\cancel{(10-y)}}{5\cancel{(10-y)}}=-\frac{11}{5}

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Answer:

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Step-by-step explanation:

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If f(x) = -3x - 5 and g(x) = 4x - 2, find (f+ g)(x).
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