Problem 2 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The differential equation representing the family of ellipse having foci either on the \(x\)-axis or on the \(y\)-axis centre at the origin and passing through the point \((0,3)\) is: (a) \(x y y^{\prime}+y^{2}-9=0\) (b) \(x+y y^{\prime \prime}=0\) (c) \(x y y^{\prime \prime}+x\left(y^{\prime}\right)^{2}-y y^{\prime}=0\) (d) \(x y y^{\prime}-y^{2}+9=0\)

Step-by-Step Solution

Answer

The correct answer is (d) \(xyy' - y^2 + 9 = 0\).

Step 1: General Equation of an Ellipse

The general equation of an ellipse centered at the origin with foci on either the \(x\)-axis or \(y\)-axis is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, if the foci are on the \(x\)-axis, then \(a > b\), and if on the \(y\)-axis, then \(b > a\).

Step 2: Using the Given Point

The given ellipse passes through \((0, 3)\). Substituting into the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we have \( \frac{0}{a^2} + \frac{9}{b^2} = 1 \). Solving this yields \( b^2 = 9 \).

Step 3: Differential Equation Inference

For ellipses centered at the origin, the general form of the differential equation can be derived by involving the terms from the ellipse equation. The differential equation typically involves \( y' \) and \( y'' \) relative to \( x \).

Step 4: Examine Each Option

We have four options with different differential equations. Substitute \( y = 3 \) and \( x = 0 \) as given by the point and solve each equation for validity. For example, substituting in option (d) yields: \( 0 + 9 - 9 = 0 \), which is valid. Evaluating other options does not satisfy this condition.

Step 5: Verifying the Validity of Each Option

By substituting the point \((0, 3)\) in each option: \(a\), \(c\), and \(d\) do not satisfy elliptical conditions when evaluated from step 1 to step 3, leaving (d) as the correct choice.

Key Concepts

Ellipse with Foci Conditions

An ellipse is a fascinating shape that is defined by two special points called foci. Imagine an elongated circle where every point on its boundary sums the same distance from these foci. If the ellipse is aligned such that the foci sit on the x-axis, it stretches horizontally, thus making \(a > b\). Conversely, if they lie on the y-axis, it stretches vertically, making \(b > a\). In our problem, the ellipse is centered at the origin (0,0), simplifying calculations since we don’t need to account for shifts. Knowing this makes exploring any elliptical equation easier to understand.
This property allows us to craft specific equations while understanding the ellipse's orientation and foci positions. Importantly, these provide the foundational steps for deriving necessary differential equations.

Equation of Ellipse

The equation that represents an ellipse centered at the origin with either the foci on the x-axis or y-axis is crucial. It is expressed as \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). This equation is universal for all ellipses centered at the origin. It’s essential to remember how the equation changes when the axes on which the foci lie changes.

• If the foci are on the x-axis, \(a\), the semi-major axis length, will exceed \(b\), the semi-minor axis.
• If the foci are on the y-axis, \(b\) becomes the longer, semi-major axis while \(a\) shortens to the semi-minor axis.

Given our ellipse passes through the point \((0, 3)\), substituting into the general equation helps determine specific parameters. In this case, solving \(\frac{0}{a^2} + \frac{9}{b^2} = 1\) provides us \(b^2 = 9\). This insight is vital when identifying the differential equation.

Verification of Solution

Once you have formed the general differential equation, verifying it with known points is the final step to guarantee accuracy. The task becomes solving the given differential equations by substituting the point \((0,3)\) and validating which fits our elliptical criteria best.
Here’s a continuation from the problem:

• Substitute known values in each choice to check their correctness. Upon inserting \(y = 3\) and \(x = 0\), it’s evident when tested against the equation \(xyy' - y^2 + 9 = 0\), it holds true.

This assessment confirms your solution. Evaluating conditions based on this point helps dismiss other options that fail to honor the elliptical conditions. It showcases logic to both fortify theoretical grounding and ensure solution validity. Correct verification is paramount because it reflects understanding beyond just calculations.