A normal at a point P of the curve meets x-axis at A and tangent at P meets y-axis at B. If the line AB is bisected at the ordinate of P, prove that the curve is a hyperbola and find its eccentricity

To tackle this problem, we need to analyze the geometric properties of the curve and the relationships between the points involved. We are given a curve with a point P, where the normal at P intersects the x-axis at point A, and the tangent at P intersects the y-axis at point B. The key condition is that the line segment AB is bisected at the ordinate of P. Let's break this down step by step to prove that the curve is a hyperbola and determine its eccentricity.

Understanding the Geometry

Let’s denote the coordinates of point P on the curve as (x₁, y₁). The slope of the tangent line at P can be represented as m, which is the derivative of the curve at that point, denoted as y' = f'(x₁).

Finding Points A and B

• Point A: The normal line at P has a slope of -1/m. The equation of the normal line can be written as:

y - y₁ = -1/m (x - x₁)

• Setting y = 0 to find the x-intercept (point A), we solve:

0 - y₁ = -1/m (x - x₁) ⟹ x = x₁ + my₁

Thus, A has coordinates (x₁ + my₁, 0).

• Point B: The tangent line at P has a slope of m. The equation of the tangent line is:

y - y₁ = m (x - x₁)

• Setting x = 0 to find the y-intercept (point B), we solve:

y - y₁ = m(0 - x₁) ⟹ y = y₁ - mx₁

Thus, B has coordinates (0, y₁ - mx₁).

Finding the Midpoint of AB

The midpoint M of line segment AB can be calculated using the midpoint formula:

M = ((x₁ + my₁ + 0)/2, (0 + (y₁ - mx₁))/2) = ((x₁ + my₁)/2, (y₁ - mx₁)/2).

Condition for Bisection

According to the problem, this midpoint M must lie on the ordinate of P, which means its x-coordinate must equal x₁. Therefore, we set:

(x₁ + my₁)/2 = x₁.

Solving this gives:

my₁ = x₁.

Relating to the Curve

From the relationship we derived, we can express m in terms of y₁ and x₁:

m = x₁/y₁.

Since m is the slope of the tangent, we can relate it to the derivative of the curve:

f'(x₁) = x₁/y₁.

Establishing the Hyperbola Condition

For the curve to be a hyperbola, we need to establish a relationship between x and y. We can rearrange the derivative condition:

y' = x/y ⟹ y dy = x dx.

Integrating both sides gives:

(1/2)y² = (1/2)x² + C,

which simplifies to:

x²/a² - y²/b² = 1,

indicating that the curve is indeed a hyperbola.

Finding the Eccentricity

The eccentricity (e) of a hyperbola is given by the formula:

e = √(1 + (b²/a²)).

In our case, we can express the eccentricity in terms of the parameters of the hyperbola derived from the relationship established earlier. If we assume a = 1 and b = 1 for simplicity, we find:

e = √(1 + 1) = √2.

Thus, we have shown that the curve is a hyperbola and determined its eccentricity. The key steps involved understanding the geometry of the normal and tangent lines, applying the midpoint condition, and relating the derivatives to the standard form of a hyperbola. This methodical approach not only proves the nature of the curve but also provides insight into its properties.