Final answer:
To show that the normals at the points of contact meet on the line y/y₁=x/x₁, we need to find the equations of the tangent and normal lines at a point on the ellipse x²a²+y²b²=1. By finding the slopes and using the point-slope form, we can find the equations of these lines. Finally, we show that the product of the slopes of the normal lines is -1, which proves that they intersect at right angles and meet on the line y/y₁=x/x₁.
Step-by-step explanation:
To show that the normals at the points of contact meet on the line y/y₁=x/x₁, let's start by finding the equation for the tangent line at a point (x₁, y₁) on the ellipse x²a²+y²b²=1.
The derivative of the implicit equation x²a²+y²b²=1 is: dy/dx = -x₁b²/a²y₁.
This gives us the slope of the tangent line at (x₁, y₁). We can then find the equation of the tangent line using the point-slope form: y - y₁ = (-x₁b²/a²y₁)(x - x₁).
Now, let's find the equation of the normal line at (x₁, y₁). The slope of the normal line is the negative reciprocal of the slope of the tangent line, which is a²y₁/(b²x₁). Using the point-slope form again, we get: y - y₁ = (a²y₁/(b²x₁))(x - x₁).
Finally, we need to show that the normals at the points of contact intersect at right angles. Two lines are perpendicular if and only if the product of their slopes is -1. Taking the product of the slopes of the normal lines, we have (a²y₁/(b²x₁)) * (a²y₂/(b²x₂)) = -1.
Simplifying this equation, we get: x₁x₂ = y₁y₂. This is the equation of the line y/y₁=x/x₁, which means the normals at the points of contact do indeed meet on this line.