Problem 188 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(P(3,3)\) be a point on the hyperbola, \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\). If the normal to it at \(P\) intersects the \(x\)-axis at \((9,0)\) and \(e\) is its eccentricity, then the ordered pair \(\left(a^{2}, e^{2}\right)\) is equal to: (a) \(\left(\frac{9}{2}, 3\right)\) (b) \(\left(\frac{3}{2}, 2\right)\) (c) \(\left(\frac{9}{2}, 2\right)\) (d) \((9,3)\)

Step-by-Step Solution

Answer

(a) \(\left(\frac{9}{2}, 3\right)\)

Step 1: Substitute Point in Hyperbola Equation

Substitute the point \(P(3,3)\) into the hyperbola equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) to find a relationship between \(a^2\) and \(b^2\). \(\frac{3^2}{a^2} - \frac{3^2}{b^2} = 1\). This simplifies to \(\frac{9}{a^2} - \frac{9}{b^2} = 1\).

Step 2: Differentiate Hyperbola Equation

Calculate the derivative to find the slope of the tangent at point \(P(3,3)\). Differentiating implicitly, we obtain: \(\frac{2x}{a^2} - \frac{2y}{b^2} \cdot \frac{dy}{dx} = 0\). Solve for \(\frac{dy}{dx}\), which gives us \(\frac{dy}{dx} = \frac{b^2x}{a^2y}\). Substitute \(x = 3\) and \(y = 3\): \(\frac{dy}{dx} = \frac{b^2 \cdot 3}{a^2 \cdot 3} = \frac{b^2}{a^2}\).

Step 3: Find Slope of the Normal Line

The slope of the normal line is the negative reciprocal of the tangent's slope, so it is \(-\frac{a^2}{b^2}\).

Step 4: Use Point-Slope Form of Normal Line

Use point \((3,3)\) and the derived slope \(-\frac{a^2}{b^2}\) in the point-slope form: \(y - 3 = -\frac{a^2}{b^2}(x - 3)\). Simplify this to \(y = -\frac{a^2}{b^2}x + \left(3 + \frac{3a^2}{b^2}\right)\).

Step 5: Find x-Intercept of Normal Line

The normal line intercepts the x-axis when \(y = 0\). Solve \(0 = -\frac{a^2}{b^2}x + \left(3 + \frac{3a^2}{b^2}\right)\). Set the equation \(\frac{a^2}{b^2}x = 3 + \frac{3a^2}{b^2}\) and substitute \(x = 9\), given that it intersects at \((9,0)\). This gives \(\frac{a^2}{b^2} \cdot 9 = 3 + \frac{3a^2}{b^2}\).

Step 6: Simplify and Solve for b^2 in terms of a^2

From \(\frac{9a^2}{b^2} = 3 + \frac{3a^2}{b^2}\), solve: \(6a^2 = 3b^2\) leading to \(b^2 = 2a^2\).

Step 7: Use b^2 = 2a^2 in the First Equation

Substitute \(b^2 = 2a^2\) back into the equation from Step 1: \(\frac{9}{a^2} - \frac{9}{2a^2} = 1\). This simplifies to \(\frac{18}{2a^2} - \frac{9}{2a^2} = 1\), which further simplifies to \(\frac{9}{2a^2} = 1\). Solve for \(a^2\): \(a^2 = \frac{9}{2}\).

Step 8: Calculate Eccentricity

Eccentricity of a hyperbola is given by \(e = \sqrt{1 + \frac{b^2}{a^2}}\). Using \(b^2 = 2a^2\), \(e = \sqrt{1 + 2} = \sqrt{3}\). Therefore, \(e^2 = 3\).

Step 9: Conclusion on Ordered Pair

Combine the results \(a^2 = \frac{9}{2}\) and \(e^2 = 3\). Therefore, the ordered pair \((a^2, e^2)\) is \((\frac{9}{2}, 3)\).

Key Concepts

Eccentricity

The concept of eccentricity is vital to distinguish between different conic sections such as circles, ellipses, parabolas, and hyperbolas. For hyperbolas, eccentricity (\(e\)) is always greater than 1, setting them apart from ellipses (with eccentricity less than 1) and parabolas (eccentricity equal to 1).
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Eccentricity of a hyperbola can be calculated using the formula: \[ e = \sqrt{1 + \frac{b^2}{a^2}} \] where \( a^2 \) and \( b^2 \) are the squared lengths of the semi-major and semi-minor axes, respectively. Understanding this formula helps to evaluate how 'stretched' a hyperbola is.
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In our specific exercise, once we determined that \( b^2 = 2a^2 \), we were able to find the value of eccentricity using the formula, resulting in \( e = \sqrt{3} \). This signifies that our hyperbola is significantly more elongated than a circle or ellipse, and thus, it behaves with the distinct properties of hyperbolas.

Tangents and Normals

In calculus, understanding tangents and normals gives insight into the behavior of curves at particular points.
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A tangent to a hyperbola at a point is a straight line that just touches the curve at that point and has the same instantaneous direction as the curve. The normal, on the other hand, is perpendicular to the tangent at the point of contact.
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The slope of the tangent line at any given point on the hyperbola can be found using differentiation. For the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the differentiation revealed that the slope of the tangent at point \((3,3)\) is \(\frac{b^2}{a^2}\). Thus, the slope of the normal line is simply the negative reciprocal: \(-\frac{a^2}{b^2}\).
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The exercise demonstrated how to determine both these slopes and use them effectively in the point-slope equation of the line to solve the problem where the normal intersects the x-axis.

Differentiation in Calculus

Differentiation is a key technique in calculus, particularly for finding the derivative of a function. The derivative at a particular point can tell us about the instantaneous rate of change of the function there, which helps in finding tangents and normals, and much more.
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When handling implicit differentiation, which is necessary for equations like those of hyperbolas, the goal is to differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\). In the given hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), differentiating both sides allows us to derive the slope of the tangent line at any point \((x, y)\).
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In our exercise, applying implicit differentiation provided a crucial expression for \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\), leading us to find the slope of the tangent line efficiently at the given point, and thereafter the normal line. Understanding this process is fundamental not just for solving problems with hyperbolas, but all conic sections, illustrating the power and utility of differentiation in calculus.