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enyata [817]
3 years ago
8

Find the points on the lemniscate where the tangent is horizontal. 8(x2 + y2)2 = 81(x2 − y2)

Mathematics
2 answers:
Snowcat [4.5K]3 years ago
6 0

Answer:

(\dfrac{9\sqrt{3}}{8},\dfrac{9}{8}),(-\dfrac{9\sqrt{3}}{8},\dfrac{9}{8}),(\dfrac{9\sqrt{3}}{8},-\dfrac{9}{8}),(-\dfrac{9\sqrt{3}}{8},-\dfrac{9}{8})

Step-by-step explanation:

The given equation of lemniscate is

8(x^2+y^2)^2=81(x^2-y^2)

Differentiate with respect to x.

16(x^2+y^2)(2x+2y\dfrac{dy}{dx})=81(2x-2y\dfrac{dy}{dx})

We know that \dfrac{dy}{dx} is slope of the tangent and slope of a horizontal line is 0, So, \dfrac{dy}{dx}=0.

16(x^2+y^2)(2x+2y(0))=81(2x-2y(0))

32x(x^2+y^2)=162x

x^2+y^2=\dfrac{162x}{32x}

x^2+y^2=\dfrac{81}{16}

x^2=\dfrac{81}{16}-y^2

Substitute this value in the given equation.

8(\dfrac{81}{16})^2=81(\dfrac{81}{16}-2y^2)

(\dfrac{81}{16})(\dfrac{1}{2})=\dfrac{81}{16}-2y^2

y^2=-\dfrac{81}{32}+\dfrac{81}{16}

y^2=\dfrac{81}{64}

y=\pm \sqrt{\dfrac{81}{64}}

y=\pm \dfrac{9}{8}

Substitute the value of y^2 in the given equation.

8(x^2+\dfrac{81}{64})^2=81(x^2\dfrac{81}{64})

On solving we get

x=\pm \dfrac{9\sqrt{3}}{8}

Therefore, the required points are (\dfrac{9\sqrt{3}}{8},\dfrac{9}{8}),(-\dfrac{9\sqrt{3}}{8},\dfrac{9}{8}),(\dfrac{9\sqrt{3}}{8},-\dfrac{9}{8}),(-\dfrac{9\sqrt{3}}{8},-\dfrac{9}{8}) .

slamgirl [31]3 years ago
3 0
<span>The answers to this problem are:<span>(<span>±5</span></span>√3/8,±5/8)<span>Here is the solution:
Step 1: <span><span><span>x2</span>+<span>y2</span>=<span>2516</span>[2]</span><span><span>x2</span>+<span>y2</span>=<span>2516</span>[2]</span></span>

Step 2: Substitute:<span> 
</span><span><span>8<span><span>(<span>25/16</span>)^</span>2</span>=25(<span>x^2</span>−<span>y^2</span>)
</span><span>8<span><span>(<span>25/16</span>)^</span>2</span>=25(<span>x^2</span>−<span>y^2</span>)</span></span>
</span><span>x^2</span>−<span>y^2</span>=<span>25/32</span><span>.

Add [2] and [3]:<span> 
</span><span>2<span>x^2</span>=<span>75/32
</span><span>x^2</span>=<span>75/74</span></span>
<span>x=±5</span></span>√3/8<span>
Substitute into [2]:<span> 
</span><span><span>75/64</span>+<span>y^2</span>=<span>50/32
</span><span>y^2</span>=<span>25/64</span></span>
<span>y=±<span>5/8</span></span>

</span>
</span>
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_____

The one rule that cannot be violated in algebra is that <em>you must do the same thing to both sides of the equation</em>.

_____

Your "solution" so far has a couple of errors. The first is that you have apparently multiplied all of the numbers by 1000. Unfortunately, when you multiply a denominator by 1000, it is the same as dividing by 1000. So, you have multiplied the left side by 1000, multiplied one term on the right by 1000 and divided another term on the right by 1000. This turns the equation into something different than what you started with, and will give a wrong answer.

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