Problem 22 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(P(a \sec \theta, b \tan \theta)\) and \(Q(a \sec \phi, b \tan \phi)\), where \(\theta+\phi=\frac{\pi}{2}\), be two points on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 .\) If \((h, k)\) is the point of intersection of the normals at \(P\) and \(Q\), then \(k\) is equal to (A) \(\frac{a^{2}+b^{2}}{a}\) (B) \(-\left(\frac{a^{2}+b^{2}}{a}\right)\) (C) \(\frac{a^{2}+b^{2}}{b}\) (D) \(-\left(\frac{a^{2}+b^{2}}{b}\right)\)

Step-by-Step Solution

Answer

\(k = \frac{a^2 + b^2}{b}\) (Answer C)

Step 1: Substitute Point Coordinates

The coordinates of point \(P\) are \( (a\sec\theta, b\tan\theta) \) and the coordinates for point \(Q\) are \( (a\sec\phi, b\tan\phi) \). The relation \( \theta + \phi = \frac{\pi}{2} \) is vital for simplifying calculations.

Step 2: Verify Points on the Hyperbola

Substitute the points into the hyperbola equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). For point \(P\), we have \( x = a\sec\theta\) and \( y = b\tan\theta \). Substitute these:\[\frac{(a\sec\theta)^2}{a^2} - \frac{(b\tan\theta)^2}{b^2} = \sec^2\theta - \tan^2\theta = 1\]This is true. Similarly, verify for \(Q\):\[\frac{(a\sec\phi)^2}{a^2} - \frac{(b\tan\phi)^2}{b^2} = \sec^2\phi - \tan^2\phi = 1\]Both points satisfy the hyperbola equation.

Step 3: Find Normal Equation

The equation of the normal at any point on the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is:\[a\frac{x}{a^2} X + b\frac{y}{b^2} Y = a^2 + b^2\]Convert this into: \[ x \cos \theta - y \sin \theta = a \sec \theta \cdot a^2 + b^2 \].Similarly, for \(Q\):\[ x \cos \phi - y \sin \phi = a \sec \phi \cdot a^2 + b^2 \].

Step 4: Solve System of Equations

Now solve the system:\[\begin{align*} x \cos \theta - y \sin \theta &= a^2 \sec \theta - b^2 \sec \theta \ x \cos \phi - y \sin \phi &= a^2 \sec \phi - b^2 \sec \phi\end{align*}\]Because \( \theta + \phi = \frac{\pi}{2} \), we use the identity \( \cos \phi = \sin \theta \) and \( \sin \phi = \cos \theta \).

Step 5: Calculate \(k\)

Substitute the values and solve for \(y\), which corresponds to \(k\):\[y = \frac{(a^2 + b^2) (\cos \theta + \cos \phi)}{a (\cos^2 \theta + \sin^2 \theta)} = \frac{a^2 + b^2}{b}\]Thus, \(k = \frac{a^2 + b^2}{b}\).

Key Concepts

Normal to a Curve

In geometry, the normal to a curve at a given point is a line perpendicular to the tangent at that point. Think of a tangent line as just touching the curve, representing the steepest slope at the point. The normal, conversely, is the opposite of this slope, standing straight up from the curve at that point.

Understanding normals is crucial because they help us analyze how curves behave in space. When we have a point on a curve, like our hyperbola, the line normal to the curve at that point can be calculated using specific rules. For hyperbolas, the normal line equation is derived from the slope of the tangent line and involves the following form:

• For the hyperbola with equation \[ rac{x^2}{a^2} - rac{y^2}{b^2} = 1 \], the normal at any point \((x_0, y_0)\) is given by \[ x rac{x_0}{a^2} + y rac{y_0}{b^2} = a^2 + b^2 \].
• This approach uses implicit differentiation to get the slope of the tangent, and then the negative reciprocal gives the slope of the normal.

It’s these normals that we solve simultaneously to find their intersection point, yielding critical points like our target coordinate \((h, k)\). Normal lines, by identifying such intersections, can shed light on specific geometric properties of curves.

Coordinate Geometry

Coordinate geometry, or analytic geometry, is a branch of mathematics that uses algebraic equations to represent geometric principles. It lets us use numbers and equations to talk about shapes and positions in a plane, making it a powerful tool for solving complex geometry problems.

In our exercise, coordinate geometry is essential in:

• Locating Points: Points \(P\) and \(Q\) are described using trigonometric functions associated with angles \(\theta\) and \(\phi\), converted to coordinates for geometrical analysis.
• Equation of the Hyperbola: It is represented as \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]. This equation helps us calculate how points \(P\) and \(Q\) lie on this curve, using known variables centered around coordinates.
• System of Equations: By substituting geometrically-defined coordinates into algebraic equations, we solve systems simultaneously for unknowns like \(h\) and \(k\).

Coordinate geometry bridges the gap between pure geometry and algebra, enabling a deeper analysis of placement and interplay between curves and lines in space. It allows precise manipulation of traditional geometric concepts.

Trigonometric Identities

Trigonometric identities play a vital role in solving problems dealing with angles and their relationships, like this hyperbola problem. These identities are equations involving trigonometric functions true for all values of the involved variables.

In our hyperbola problem, we use trigonometric identities to relate the points \(P\) and \(Q\). Here's how it works:

• The given condition \(\theta + \phi = \frac{\pi}{2}\) lets us use fundamental identities such as \( \cos \phi = \sin \theta \) and \( \sin \phi = \cos \theta \). These express one angle in terms of another, simplifying otherwise complex algebra to manageable pieces.
• Identities such as \(\sec^2 \theta - \tan^2 \theta = 1\) confirm that points are on the hyperbola, crucial for validating our solutions.

Trigonometric identities help streamline calculations by reducing the number of independent variables, letting us solve more efficiently. When applied correctly, they simplify integrations, derivatives, and verifications massively, enabling deeper understanding of geometry problems involving curves like a hyperbola.