Problem 214 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The tangent at an extremity (in the first quadrant) of latus rectum of the hyperbola \(\frac{x^{2}}{4}-\frac{y^{2}}{5}=1\), meet \(x\)-axis and \(y\)-axis at \(\mathrm{A}\) and \(\mathrm{B}\) respectively. Then \((\mathrm{OA})^{2}-(\mathrm{OB})^{2}\), where \(\mathrm{O}\) is the origin, equals: (a) \(-\frac{20}{9}\) (b) \(\frac{16}{9}\) (c) 4 (d) \(-\frac{4}{3}\)

Step-by-Step Solution

Answer

\((OA)^2 - (OB)^2 = 0\).

Step 1: Identify Equation of the Latus Rectum

For the hyperbola \(\frac{x^2}{4} - \frac{y^2}{5} = 1\), the length of the semi-latus rectum is \(\frac{b^2}{a}\) where \(a^2 = 4\) and \(b^2 = 5\). First, calculate the semi-latus rectum: \(\frac{b^2}{a} = \frac{5}{2}\). Hence, the complete latus rectum is of length \(5\).

Step 2: Find Coordinates of Latus Rectum

The equation of the hyperbola allows us to take \((a, 0)\) and find one endpoint of the latus rectum. The coordinates for points on the latus rectum above the \(x\)-axis can be computed as \((a, \pm \frac{b^2}{a} ) = (2, \pm \frac{5}{2})\). Consider now the point in the first quadrant: \((2, \frac{5}{2})\).

Step 3: Find Slope of Tangent at the Point

The equation of the tangent to the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) at a point \((x_1, y_1)\) is \(\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1\). Set \((x_1, y_1) = (2, \frac{5}{2})\) then plug into formula: \(\frac{2x}{4} - \frac{(5/2)y}{5} = 1\) gives \(\frac{x}{2} - \frac{y}{2} = 1\). Rearranging gives \(x - y = 2\).

Step 4: Find Intersection with Axes

To find where it meets the \(x\)-axis, set \(y = 0\) in \(x - y = 2\): \(x = 2\), so \(A = (2, 0)\). For the \(y\)-axis, set \(x = 0\) to get \(-y = 2\), so \(y = -2\), giving \(B = (0, -2)\).

Step 5: Calculate OA and OB

Find \(OA\) as the distance from \((0, 0)\) to \((2, 0)\), which is \(\sqrt{(2-0)^2 + (0-0)^2} = 2\). Similarly, \(OB\) is the distance to \((0, -2)\), which is \(\sqrt{(0-0)^2 + (-2-0)^2} = 2\).

Step 6: Compute \((OA)^2 - (OB)^2\)

Substitute the distances calculated: \((OA)^2 = 4\) and \((OB)^2 = 4\). Then \((OA)^2 - (OB)^2 = 4 - 4 = 0\).

Key Concepts

Latus Rectum

The term "latus rectum" refers to a particular line segment related to conic sections such as ellipses, parabolas, and hyperbolas. In the context of a hyperbola, the latus rectum is a line parallel to one of the axes of symmetry, passing through a focus of the hyperbola. This line lies entirely within the hyperbola and is perpendicular to the transverse axis.

For any hyperbola represented by \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the length of the semi-latus rectum is given by the formula \(\frac{b^2}{a}\). For our specific exercise, where \(a^2 = 4\) and \(b^2 = 5\), we calculate the semi-latus rectum as \(\frac{5}{2}\) and hence the full latus rectum as \(5\). This defines the line segment that intersects the hyperbola at significant, symmetrical points with respect to its foci.

The latus rectum is crucial in defining the geometric properties of hyperbolas, influencing both the overall shape and symmetrical properties, especially regarding tangents and directrices.

Tangent to Hyperbola

A tangent line to a hyperbola is a straight line that touches the hyperbola at exactly one point, called the point of tangency. At this point, the tangent line is perpendicular to the radius connecting the point of tangency to the center of the hyperbola.

The general equation for the tangent to a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) at a point \((x_1, y_1)\) is: \(\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1\). In our exercise, we apply this at the point \((2, \frac{5}{2})\), resulting in the tangent line equation \(x - y = 2\).

Understanding tangents is vital because they provide insights into the slopes and geometric orientation of the hyperbola at any given point. Tangents are useful for constructing perspectives and analyzing light paths, among other applications in higher-level calculus and geometry.

Distance Formula

The distance formula is a mathematical tool used to calculate the distance between two points in a plane. If you have two points \((x_1, y_1)\) and \((x_2, y_2)\), the formula is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
If we apply this to the points in our solution, such as from the origin \((0, 0)\) to the point \((2, 0)\) on the x-axis, we have:
\[ OA = \sqrt{(2 - 0)^2 + (0 - 0)^2} = 2 \]
The same calculation applies for point \((0, -2)\) on the y-axis:
\[ OB = \sqrt{(0 - 0)^2 + (-2 - 0)^2} = 2 \]
The distance formula is essential in Euclidean geometry, providing a straightforward means of determining the separation between points on a plane, influencing countless computations in analytic geometry.

Intersection with Axes

The intersection points of a line with the coordinate axes are crucial in analyzing geometric properties, allowing us to visualize where a curve or line crosses the x-axis or y-axis.

For a line, such as our tangent \(x - y = 2\), we find intersections by setting either \(x = 0\) or \(y = 0\):

• Setting \(y = 0\), we find intersection with the x-axis at \((2, 0)\).
• Setting \(x = 0\), we find intersection with the y-axis at \((0, -2)\).

Intersections help define segment lengths on the axes, pivotal in calculating distances or differences, like the squared distances in our exercise. They also help in sketching the graph of the equation visually and can guide more complex analytical techniques in calculus, such as finding critical damping values or adjusting graphs of related functions.