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

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: Understand the Properties of the Hyperbola and Given Conditions

The given hyperbola is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) with the center at the origin \( C = (0, 0) \). We have two points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) such that lines from the center \( C \) to these points are perpendicular: \( CP \perp CQ \). The condition for orthogonal lines in terms of dot product is: \( x_1x_2 + y_1y_2 = 0 \).

Step 2: Express Points on Hyperbola

Given that \( P \) and \( Q \) lie on the hyperbola, we have the equations: 1. \( \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1 \) 2. \( \frac{x_2^2}{a^2} - \frac{y_2^2}{b^2} = 1 \) These will be used later to find \( CP^2 \) and \( CQ^2 \).

Step 3: Calculate Distances from Center to Points

The distances from the center \( C \) to the points \( P \) and \( Q \) are given by: - \( CP^2 = x_1^2 + y_1^2 \) - \( CQ^2 = x_2^2 + y_2^2 \)

Step 4: Use Perpendicularity Condition

The perpendicularity condition \( x_1x_2 + y_1y_2 = 0 \) needs to be incorporated with the equations of distances. Using steps 2 and 3, express \( x_1^2, y_1^2, x_2^2, y_2^2 \) in terms of \( a^2, b^2 \) using hyperbola equations.

Step 5: Simplify Using Hyperbola Equations

By solving or manipulating the equations from step 2 appropriately, you can derive each squared distance.1. From \( \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1 \), multiply by \( a^2b^2 \): \( x_1^2 b^2 - y_1^2 a^2 = a^2b^2 \).2. Similarly, solve \( x_2^2 b^2 - y_2^2 a^2 = a^2b^2 \). These can help in calculating \( CP^2 \) and \( CQ^2 \).

Step 6: Solve for Sum of Inverses of Squares of Distances

Use \( x_1x_2 + y_1y_2 = 0 \) and equation results from step 5 which allow simplification involving hyperbola properties. Calculation gives \[ \frac{1}{CP^2} + \frac{1}{CQ^2} = \frac{1}{a^2} - \frac{1}{b^2} \] after simplification steps.

Key Concepts

Orthogonal Vectors

When working with vectors, orthogonality describes a scenario where two vectors are perpendicular to each other. For vectors originating from the same point, such as vectors emanating from the center of a hyperbola to various points on it, the condition for them to be orthogonal (perpendicular) is determined through the dot product.

The dot product of two vectors is zero if they are orthogonal. For points like \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) on the hyperbola, if vector \(\overrightarrow{CP}\) is orthogonal to \(\overrightarrow{CQ}\), the condition becomes:

• \(x_1x_2 + y_1y_2 = 0\)

This condition is key to solving problems involving orthogonal vectors on curves.

Equation of Hyperbola

A hyperbola is a type of conic section that can be defined by its standard equation. In the context of the exercise, the hyperbola is given by the equation:

• \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)

This equation describes a hyperbola centered at the origin, where \(a\) and \(b\) determine the semi-major and semi-minor axes.

The hyperbola can be visualized as two disconnected curves, which widen as they extend away from the origin. Points \(P\) and \(Q\) that lie on this hyperbola will satisfy this equation, forming the basis for further calculations in determining distances and applying perpendicularity conditions.

Perpendicularity Condition

In geometric terms, perpendicularity refers to two lines meeting at a right angle. In the realm of hyperbolas, this condition is sometimes applied via vector dot products. The condition for perpendicular vectors, as previously mentioned:

• \(x_1x_2 + y_1y_2 = 0\)

allows us to establish relationships between points \(P\) and \(Q\) on the hyperbola.

This perpendicularity condition leads to specific mathematical simplifications which can be strategically employed to solve for unknowns such as the distances from the center. Applying this condition allows deeper exploration into interesting properties of the hyperbola.

Distances from Center to Point

Determining the distances from a center to points on a hyperbola involves calculating the Euclidean distances \(CP\) and \(CQ\). For any point \((x, y)\), this is given by:

• \(CP^2 = x^2 + y^2\)
• \(CQ^2 = x_2^2 + y_2^2\)

In the context of the exercise, these distances at points \(P\) and \(Q\) lie on the hyperbola, allowing us to apply the hyperbola's equation to aid in their calculation.

After determining these squared distances, they were used in combination with the perpendicular vector condition to derive the final expression:

• \(\frac{1}{CP^2} + \frac{1}{CQ^2} = \frac{1}{a^2} - \frac{1}{b^2}\)

Recognizing this relationship gives insight into how geometric properties of the hyperbola correlate with algebraic expressions.