How to Find the Coordinates of the Ends of the Diameter of the Ellipse Conjugate to a Given Line

The problem involves identifying the coordinates of the ends of the diameter of an ellipse that is conjugate to a given line, specifically the ellipse defined by the equation 16x^2 25y^2 1000 and the line 5y 4x. This guide walks through every step of the process to solve this problem, ensuring clarity and precision at each stage.

Step 1: Transforming the Ellipse Equation into Standard Form

The first step is to convert the equation of the ellipse into its standard form. This conversion simplifies the process of identifying key properties of the ellipse.

Start by dividing the entire equation by 1000:

frac{16x^2}{1000} frac{25y^2}{1000} 1

This simplifies to:

frac{x^2}{62.5} - frac{y^2}{40} 1

Step 2: Identifying Key Properties of the Ellipse

The standard form of the ellipse is:

frac{x^2}{a^2} - frac{y^2}{b^2} 1

From this, we can identify the lengths of the semi-major and semi-minor axes:

a^2 62.5, so a sqrt{62.5} approx 7.91 b^2 40, so b sqrt{40} approx 6.32

Step 3: Determining the Slope of the Conjugate Diameter

The given line 5y 4x can be rewritten in slope-intercept form as:

y frac{4}{5}x

Thus, the slope of this line, m, is:

m frac{4}{5}

Step 4: Calculating the Slope of the Conjugate Diameter

The slope of the conjugate diameter is the negative reciprocal of the given line's slope:

m_{conjugate} - frac{5}{4}

Step 5: Formulating the Equation of the Conjugate Diameter

The equation of the diameter passing through the center of the ellipse at (0, 0) with the calculated slope:

y - frac{5}{4}x

Step 6: Solving for Intersection Points

Substitute y -frac{5}{4}x into the ellipse equation:

frac{x^2}{62.5} - frac{left(-frac{5}{4}xright)^2}{40} 1

After simplification:

frac{x^2}{62.5} - frac{25}{16} cdot frac{x^2}{40} 1

Further simplification results in:

frac{16^2}{80000}x^2 - frac{5^2}{80000}x^2 1

Multiplying both sides by 80000:

256x^2 - 625x^2 80000

Combining terms:

-369x^2 80000

Thus:

x^2 frac{80000}{369} approx 215.8024691

So, x approx pm sqrt{215.8024691} approx pm 14.70

Step 7: Finding the Corresponding y Values

Substituting the x values back into the conjugate diameter equation gives the y values:

For x approx 14.70, y approx -11.68 (since y - frac{5}{4}x)

For x approx -14.70, y approx 11.68 (since y - frac{5}{4}x)

Final Coordinates of the Ends of the Diameter

The coordinates of the ends of the diameter are approximately:

(14.70, -11.68) and (-14.70, 11.68)