Question

Find the evolutes of x^2/a^2 + y^2/b^2=1

Answer 1

Answer:

(y - y₀)a²y₀ + (x - x₀)/x₀b² = 0

Step-by-step explanation:

differentiating the Equation x²/a² - y²/b² = 1 with respect to x .

2x/a² - 2y/b² * dy/dx = 0

2x/a² = 2y/b² * dy/dx

xb²/ya² = dy/dx

Now, the slope at point (X₀ , Y₀) is;

dy/dx_{x₀ , y₀} = x₀b²/y₀a²

equation of tangent is;

y - y₀ = dy/dx(x - x₀)

y - y₀ = X₀b²/y₀a²(x - x₀)

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a²(yy₀ - y₀²) = b²(xx₀ - x₀²)

yy₀/b² - y₀²/b² = xx₀a² - x₀²/a²

yy₀/b² - xx₀/a² = y₀²/b² - x₀²/a²

yy₀/b² - xx₀/a² = -(x₀²/a² - y₀b²)

since, (x₀ , y₀) is point in the curve

So, from the given Equation x₀a² - y₀/b² = 1

Hence, the equation of tangent is: xx₀a² - yy₀/b² = 1

now, the slope of normal = -1/dy/dx_{x₀ , y₀} = y₀a²/x₀b²

Now, the equation is;

y - y₀ = -1/(dy/dx) * (x - x₀)

y - y₀ = -(y₀a²/x₀b²) * (x - x₀)

(y - y₀)/a²y₀ = -(x - x₀)/x₀b²

(y - y₀)a²y₀ + (x - x₀)/x₀b² = 0