Question

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

Answer 1

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

Answer 2

The answers to this problem are:(±5√3/8,±5/8)Here is the solution:
Step 1: x2+y2=2516[2]x2+y2=2516[2]

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

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