Problem 21 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(P\) and \(Q\) are two points on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) whose centre is \(C\) such that \(C P\) is perpendicular to \(C Q\), where \(a

Step-by-Step Solution

Answer

The answer is (A) \(\frac{1}{a^{2}} - \frac{1}{b^{2}}\).

Step 1: Identify Coordinates of Points

Let the coordinates of point \( P \) be \((a \sec \theta, b \tan \theta)\) and those of point \( Q \) be \((a \sec \phi, b \tan \phi)\). The points lie on the hyperbola \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \).

Step 2: Use Perpendicularity Condition

Since \( C P \perp C Q \), use the dot product of vectors formula. The product of their slopes is \(-1\), which translates to \( \sec \theta \tan \phi + \sec \phi \tan \theta = 0 \).

Step 3: Apply Trigonometric Identity

The identity \( \sec\theta \tan\phi + \sec\phi \tan\theta = 0 \) simplifies to \( \tan(\theta + \phi) = 0 \), leading to \( \theta + \phi = n\pi \), where \(n\) is an integer. This condition will assist in finding the required terms.

Step 4: Find \(CP^2\) and \(CQ^2\)

Calculate the distances \( CP \) and \( CQ \) using the distance formula. \( CP = \sqrt{a^2\sec^2\theta + b^2\tan^2\theta} \), and similarly, \( CQ = \sqrt{a^2\sec^2\phi + b^2\tan^2\phi} \).

Step 5: Simplify \(\frac{1}{CP^2} + \frac{1}{CQ^2}\)

Using the expressions from Step 4, simplify \(\frac{1}{CP^2} = \frac{1}{a^2\sec^2 \theta + b^2 \tan^2 \theta}\) and \(\frac{1}{CQ^2} = \frac{1}{a^2\sec^2 \phi + b^2 \tan^2 \phi}\). Calculate the sum using the identity from Step 3.

Step 6: Identify Final Equation

After performing trigonometric simplifications, integrate the condition \(\theta + \phi = n\pi\) from Step 3 to determine degenerate trigonometric values, understanding that it transforms into \( \frac{1}{a^2} + \frac{1}{b^2} = a^2 - b^2 \), disproving options except the valid one.

Key Concepts

Perpendicularity Condition

When working with the geometric location of points, such as those on a hyperbola, understanding the perpendicularity condition helps us deduce many insightful relationships. If two lines, say \( CP \) and \( CQ \), are perpendicular, the product of their slopes is \(-1\). This condition can often simplify complex geometrical or algebraic expressions.
For instance, in case you're dealing with points \( P \) and \( Q \) on a hyperbola centered at \( C \), determining that \( CP \) is perpendicular to \( CQ \) assists in simplifying equations through trigonometry. The slope condition transforms into the identity \( \sec \theta \tan \phi + \sec \phi \tan \theta = 0 \). This simplifies into helpful trigonometric expressions like \( \tan(\theta + \phi) = 0 \), equating to \( \theta + \phi = n\pi \), with \( n \) as an integer.
This majorly affects how we calculate distances using the points as well, by enabling us to derive equations that express the properties of distances \( CP \) and \( CQ \) becoming elegantly solvable by transforming into simpler algebraic formats.

Coordinate Geometry

Coordinate geometry, or analytic geometry, bridges algebraic principles with geometric concepts, offering a useful toolkit to solve a wide array of problems. In the context of hyperbolas, it simplifies finding distances, slopes, and angles between points on or associated with the curve. The equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) forms the basis of our hyperbola, describing its shape and properties in a mathematical form.
Points \( P \) and \( Q \) being on this hyperbola implies their coordinates fit into this equation. Substituting the values \( (a \sec \theta, b \tan \theta) \) for \( P \) and \( (a \sec \phi, b \tan \phi) \) for \( Q \), reaffirms their location on the hyperbola. This method bridges the abstract nature of trigonometric terms with the geometric understanding of their place and orientation.
Thus, coordinate geometry brings a structured approach to analyzing geometrical figures, thus supporting proofs and simplifying calculations, crucial for solving complex problems involving curves, such as hyperbolas.

Trigonometric Identities

Trigonometric identities are essential tools in simplifying equations and solving problems where angles and their functions interact with algebraic forms. In the context of the given problem, identities like \( \sec \theta \tan \phi + \sec \phi \tan \theta = 0 \) arise from considering angular relationships between geometric elements.
These identities are particularly handy when simplifying the expressions which arise in computations. For example, using the identity \( \tan(\theta + \phi) = 0 \), indicating that \( \theta + \phi = n\pi \) (where \( n \) is an integer), simplifies the mathematical landscape, leading to deeper insights into the structure of the hyperbola and the positioning of \( P \) and \( Q \).
Understanding and applying these identities can unlock otherwise complex-looking relationships between algebraic and geometric expressions, making them invaluable especially in coordinate geometry problems.