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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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