Problem 128 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\mathrm{L}_{1}\) be the length of the common chord of the curves \(\mathrm{x}^{2}+\mathrm{y}^{2}=9\) and \(\mathrm{y}^{2}=8 \mathrm{x}\), and \(\mathrm{L}_{2}\) be the length of the latus rectum of \(y^{2}=8 x\), then: (a) \(\mathrm{L}_{1}>\mathrm{L}_{2}\) (b) \(\mathrm{L}_{1}=\mathrm{L}_{2}\) (c) \(\mathrm{L}_{1}<\mathrm{L}_{2}\) (d) \(\frac{L_{1}}{L_{2}}=\sqrt{2}\)

Step-by-Step Solution

Answer

Cannot directly solve as presented without further clarification on L1.

Step 1: Identify the Intersection Points

First, we need to find the intersection points of the two curves. The common chord is the line segment joining these points. The equations given are \(x^2 + y^2 = 9\) and \(y^2 = 8x\). To find the intersection points, substitute \(y^2 = 8x\) into the equation \(x^2 + y^2 = 9\):\[x^2 + 8x = 9\]

Step 2: Solve the Intersection Equation

Rearrange the equation from Step 1 into a standard quadratic form:\[ x^2 + 8x - 9 = 0. \]Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 8\), and \(c = -9\), compute the roots:\[ x = \frac{-8 \pm \sqrt{64 + 36}}{2} \Rightarrow x = \frac{-8 \pm \sqrt{100}}{2} \Rightarrow x = \frac{-8 \pm 10}{2}. \]This gives \(x = 1\) and \(x = -9\).

Step 3: Find Corresponding y-values

Find the \(y\)-coordinates corresponding to each \(x\) value. When \(x = 1\), substituting in \(y^2 = 8x\) gives:\[ y^2 = 8(1) = 8 \Rightarrow y = \pm \sqrt{8} = \pm 2\sqrt{2}. \]For \(x = -9\), \(y^2 = 8(-9) = -72\), which is not possible in the real numbers, so only \(x = 1\) gives real points of intersection.

Step 4: Calculate L1 - Length of Common Chord

Since there is only one real intersection point at \((1, \pm 2\sqrt{2})\), we reconsider the interpretation. It's known that the equation \(x^2 + y^2 = 9\) represents a circle, and generally, a chord cannot exist if the curves only intersect at one real point. By proceeding with this, it's shown that the expected form is misunderstood: we are looking at situations, not in general application but specific instances which will not directly provide the common chord using aforementioned methods.

Step 5: Determine L2 - Length of Latus Rectum

The given parabola is \(y^2 = 8x\). The formula for the length of the latus rectum for \(y^2 = 4ax\) is \(4a\). Here, \(4a = 8\), so \(a = 2\). Thus, the length of the latus rectum, \(L_2\), is \(4 \times 2 = 8\).

Step 6: Compare L1 and L2

We now compare \(L_1\) and \(L_2\). Since the derivation based on real intersection was incorrectly formulated, using common mathematical analysis in this theoretical approach must suffice before concluding exact computation. Normally, error checking would continue as actual implementation could support more detail.

Key Concepts

Conic Sections

Conic sections are fascinating shapes that are formed by the intersection of a plane with a double-napped cone. They include shapes such as circles, ellipses, parabolas, and hyperbolas. Each of these shapes has unique properties and equations that define them.
For example, a circle has an equation of the form \(x^2 + y^2 = r^2\), where \(r\) is the radius. This tells us that all points on the circle are equidistant from the center.
The parabola, on the other hand, can be represented by equations like \(y^2 = 4ax\). This indicates that a parabola has a focus point and a directrix such that each point on the curve is equidistant from the focus and the directrix.
Conic sections are common in algebra, and understanding them is crucial for problems like the intersection of curves, where knowing the nature of the curves helps in visualizing and solving problems.

Intersection of Curves

When exploring the intersection of curves, we are essentially looking for the set of points that satisfy both of the given curve equations simultaneously.
In the original exercise, we have two curves: a circle with equation \(x^2 + y^2 = 9\) and a parabola given by \(y^2 = 8x\). To find their intersection points, we substitute the expression for \(y^2\) from the parabola into the circle's equation, which simplifies to a quadratic equation in terms of \(x\).
This quadratic equation can be solved using methods such as factoring, completing the square, or the quadratic formula. Solving it will give us the \(x\) values of the intersection points. The corresponding \(y\) values can be found by substituting back into one of the original equations.

• The number of real solutions determines how many real intersection points exist.
• In our specific case, if there's only one real solution, it suggests the curves touch at a single point, perhaps tangentially.

Understanding the intersections of curves is essential for determining distances between points on the curves, like the common chord length in this problem.

Quadratic Equations

Quadratic equations are fundamental in algebra, expressed in the standard form \(ax^2 + bx + c = 0\). These equations can have zero, one, or two real solutions, depending on the discriminant \(b^2 - 4ac\).
The solutions can be found using the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
In the exercise, we derived a quadratic equation \(x^2 + 8x - 9 = 0\) while finding the intersection of the circle and parabola. The discriminant, \(b^2 - 4ac\), guides us in understanding how many real solutions there are:

• If it's positive, there are two distinct solutions.
• If it's zero, there is exactly one solution, indicating that the curves are tangent at a point.
• If it's negative, there are no real solutions, meaning the curves do not intersect in the real plane.

Quadratics also help compute key features in geometry, such as the length of the latus rectum in conic sections. Calculations involving quadratic equations highlight the interconnected concepts in higher mathematics.