Problem 148 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the point \(\mathrm{P}\) on the curve, \(4 x^{2}+5 y^{2}=20\) is farthest from the point \(\mathrm{Q}(0,-4)\), then \(\mathrm{PQ}^{2}\) is equals to : (a) 36 (b) 48 (c) 21 (d) 29

Step-by-Step Solution

Answer

The maximum value of \(PQ^2\) is 36.

Step 1: Identify the Curve Equation

The given curve is an ellipse and its equation is \(4x^2 + 5y^2 = 20\). We can rewrite it in standard form by dividing the whole equation by 20: \(\frac{x^2}{5} + \frac{y^2}{4} = 1\). This represents an ellipse centered at the origin.

Step 2: Distance Formula Setup

We want to find the point on the ellipse that is farthest from the point \(Q(0, -4)\). The squared distance between a point \((x, y)\) on the ellipse and \(Q(0, -4)\) is \(PQ^2 = (x - 0)^2 + (y + 4)^2 = x^2 + (y + 4)^2\).

Step 3: Substitute and Simplify the Distance

Substitute \(y\) from the ellipse equation. From \(4x^2 + 5y^2 = 20\), express \(y^2 = \frac{20 - 4x^2}{5}\). Subsequently, \((y + 4)^2 = y^2 + 8y + 16\). We will substitute \(y^2\) and analyze to maximize \(PQ^2\).

Step 4: Utilize Lagrange Multipliers

To maximize \(PQ^2 = x^2 + y^2 + 8y + 16\) under the constraint \(4x^2 + 5y^2 = 20\), apply Lagrange multipliers. Set \(abla f = abla g\lambda\), where \(f(x, y) = x^2 + y^2 + 8y + 16\) and \(g(x, y) = 4x^2 + 5y^2 - 20\).

Step 5: Solve the System of Equations

Solve the system of equations given by Lagrange multipliers. For \(x\), \(2x = 8x\lambda\), leading to \(\lambda = \frac{1}{4}\). For \(y\), \(2y + 8 = 10y\lambda\), simplifies to solve for \(y\). Substitute back to find corresponding \(x\), ensuring it satisfies the original ellipse equation.

Step 6: Calculate Maximum Distance squared

Using the solutions \((x, y)\) obtained from Lagrange multipliers, calculate \(PQ^2\). Substitute these values back into the \(PQ^2 = x^2 + (y+4)^2\) to find the maximum distance squared that corresponds to when the point \(P\) is farthest from \(Q\).

Key Concepts

Distance formula

The distance formula represents a crucial mathematical tool for calculating the distance between two points in a plane. It derives from the Pythagorean theorem, used to calculate lengths of sides in right-angled triangles. For any two points, \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the formula is:\[D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]In this exercise, we are interested in the distance from a point on an ellipse to another fixed point, specifically point \(Q(0, -4)\). Here, the formula is used slightly differently; we calculate the squared distance to simplify the algebra within this optimization problem. Thus, the squared distance is represented as:\[PQ^2 = (x - 0)^2 + (y + 4)^2 \]This adjustment avoids the complication of working with square roots, making it easier to differentiate and work with the equations during optimization processes.

Lagrange multipliers

Lagrange multipliers provide a strategy for finding the local maxima or minima of a function subject to equality constraints. This powerful mathematical method is particularly useful in multivariable calculus to leverage when dealing with boundaries or conditions. Consider a function \(f(x, y)\) you want to maximize or minimize subject to a constraint \(g(x, y) = c\).The method involves introducing a new variable called the Lagrange multiplier (\(\lambda\)) and setting the gradients equal by forming the equation \(abla f = \lambda abla g\). Here, \(abla f\) and \(abla g\) are the gradients of the functions:\[abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\]\[abla g = \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}\right)\] In our problem, \(f(x, y) = x^2 + y^2 + 8y + 16\) expresses the distance squared, and \(g(x, y) = 4x^2 + 5y^2 - 20\) is the ellipse equation. Solving these allows us to find the necessary \(x\) and \(y\) values that provide the maximal distance subject to the constraint.

Conic sections

Conic sections are curves derived from the intersection of a plane and a double-napped cone, encompassing circles, ellipses, parabolas, and hyperbolas. In this context, an ellipse represents the set of all points such that the sum of the distances from two fixed points (foci) is constant.An ellipse has its own standard form equation:\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]where \(a\) and \(b\) are distances from the center to the ellipse along the x-axis and y-axis, respectively. In the problem at hand, the original equation \(4x^2 + 5y^2 = 20\) transformed into its standard form reveals the axes and the orientation of the ellipse:\[\frac{x^2}{5} + \frac{y^2}{4} = 1\]Understanding this form is essential to identifying the geometry of the ellipse, which is critical in applying optimization techniques such as Lagrange multipliers.

Optimization problem

Optimization in mathematics involves finding an extreme value, either the minimum or maximum, for a particular function. It is widely applicable in various fields like economics, engineering, and physics, where it is imperative to find optimal solutions for certain conditions. The process often involves specifying a function to be optimized and any constraints under which the function operates.In the given exercise, we are tasked with maximizing the squared distance (\(PQ^2 = x^2 + (y+4)^2\)) between a point on the ellipse and another point \(Q(0, -4)\). Subject to the constraint provided by the equation of the ellipse, the problem becomes fining points on the boundary of this ellipse that are farthest from the fixed point.By using Lagrange multipliers, and employing calculus-based strategies, the strengths of optimization theory help identify the relevant \((x, y)\) coordinates that show these extreme values.