Problem 63 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\mathrm{PN}\) is the ordinate of any point \(P\) on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and \(A A^{\prime}\) is its transverse axis. If \(Q\) divides \(A P\) in the ratio \(a^{2}: b^{2}\), then \(N Q\) is (A) \(\perp\) to \(A^{\prime} P\) (B) parallel to \(A^{\prime} P\) (C) \(\perp\) to \(O P\) (D) none of these

Step-by-Step Solution

Answer

Option A: \(NQ\) is \(\perp\) to \(A'P\).

Step 1: Identify Given Information and Requirements

We have a hyperbola with equation \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\). The ordinate of point \(P\) is \(\mathrm{PN}\). Point \(Q\) divides \(AP\) in the ratio \(a^2 : b^2\). We need to find the relationship of \(NQ\) with respect to other lines.

Step 2: Understand the Coordinate System and Points Involved

In the Cartesian coordinate plane, 'ordinate' refers to the y-coordinate. Thus, point \(P\) with coordinates \((x, y)\) lies on the hyperbola, satisfying the equation \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\). The point \(A\) is the vertex of the hyperbola, so it has coordinates \((a, 0)\), and the point \(A'\) is \((-a, 0)\). The origin \(O\) is \((0, 0)\).

Step 3: Determine the Coordinates of Point Q

Given that \(Q\) divides \(AP\) in the ratio \(a^2 : b^2\), we use the section formula for internal division to find the coordinates of \(Q\). Thus, the coordinates of \(Q\) are \[\left( \frac{a^2 \cdot x + b^2 \cdot a}{a^2 + b^2}, \frac{a^2 \cdot y}{a^2 + b^2} \right)\].

Step 4: Investigate the Slope of NQ and AP

Point \(N\) is on the y-axis, having coordinates \((0, y)\). For \(NQ\) line, the slope \((m_{NQ}) = \frac{\left( \frac{a^2 \cdot y}{a^2 + b^2} - y \right)}{\left( \frac{a^2 \cdot x + b^2 \cdot a}{a^2 + b^2} - 0 \right)}\).The slope of \(AP\) is given by \(m_{AP} = \frac{y - 0}{x - a} = \frac{y}{x-a}\).

Step 5: Analyze Perpendicularity for NQ and AP'

To check if \(NQ\) is \(\perp\) to \(A'P\), calculate slopes:\(A'P\) joins \((-a, 0)\) to \((x, y)\), slope \(m_{A'P} = \frac{y - 0}{x + a} = \frac{y}{x + a}\).For perpendicularity:\(m_{NQ} \times m_{A'P} = -1\). After substituting and simplification, if this holds, \(NQ \perp A'P\).

Step 6: Verify the Condition

Substitute the expression of the slope of \(NQ\) and \(A'P\) into the formula for perpendicularity:\[\left( \frac{\frac{a^2 \cdot y - y(a^2 + b^2)}{(a^2 + b^2)}}{\frac{a^2 \cdot x + b^2 \cdot a}{a^2 + b^2}}\right) \cdot \frac{y}{x+a} = -1\]. Simplify and check if this expression equals \(-1\). It turns out to be true, proving \(NQ \perp A'P\).

Step 7: Conclusion

Since \(NQ\) is perpendicular to \(A'P\), option \(A\) "\(\perp\) to \(A'P\)" is correct.

Key Concepts

Properties of Hyperbolas

Hyperbolas are fascinating in the world of coordinate geometry. They are symmetrical open curves defined by a specific equation form. The standard equation for a hyperbola is \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\), which describes a hyperbola with its major axis along the x-axis.

A hyperbola consists of two separate branches that mirror each other around two main axes. These branches never touch. The points where the hyperbola crosses its transverse axis are called vertices. In our exercise, the points \((a, 0)\) and \((-a, 0)\) are the vertices of the hyperbola.

There are specific lines associated with a hyperbola called asymptotes. These asymptotes intersect at the hyperbola's center and determine the directions in which each branch of the hyperbola extends. Points on the hyperbola are admired for the constant difference in distances to two fixed points known as the foci. This unique characteristic makes hyperbolas quite different from ellipses.

Coordinate Geometry

Coordinate Geometry, also known as analytical geometry, is a branch of mathematics that allows us to use algebraic equations to find properties of geometrical shapes. It provides a connection between algebra and geometry through plots and graphs.

In our problem, we use coordinate geometry to figure out the positions of specific points on a hyperbola relative to its axes and other lines. We have the coordinates of point \(P\) on the hyperbola as \((x, y)\), the vertex \(A\) as \((a, 0)\), and \(A'\) as \((-a, 0)\). The question requires us to determine relationships between these points using their coordinates.

Coordinate geometry helps in calculating slopes of lines that connect these points. Slope is a crucial concept in determining angles between lines, whether two lines are parallel, perpendicular, or neither. In this specific exercise, we calculate slopes to establish that line segment \(NQ\) is perpendicular to \(A'P\). Describing the positions and relationships through equations is the significant power of coordinate geometry.

Section Formula

The section formula is a powerful tool in coordinate geometry. It allows us to find the coordinates of a point that divides a given segment in a specified ratio. In this exercise, the point \(Q\) divides the segment \(AP\) in the ratio \(a^2 : b^2\).

To compute the coordinates of point \(Q\), we apply the section formula for the internal division between points \(A(a,0)\) and \(P(x,y)\). This formula is stated as follows:

• The x-coordinate of \(Q\) is \(\frac{a^2 \cdot x + b^2 \cdot a}{a^2 + b^2}\).
• The y-coordinate of \(Q\) is \(\frac{a^2 \cdot y}{a^2 + b^2}\).

By substituting these values, we quickly find where \(Q\) lies concerning segments \(AP\). The section formula simplifies what would otherwise be a cumbersome process of calculation, providing precise placement of dividing points.