Question

If the curves \(2 \mathrm{x}^{2}+3 \mathrm{y}^{2}=6\) and \(\mathrm{ax}^{2}+4 \mathrm{y}^{2}=4\) intersect orthogonally than a = (a) 2 (b) 1 (c) 3 (d) \(-3\)

Short answer

The value of 'a' is 1 for the curves to intersect orthogonally. Answer: \(\boxed{\text{(b) } 1}\).

Step-by-step solution

Differentiate the given curves

Differentiate both curves with respect to x: For the first curve, \(2x^2 + 3y^2 = 6\), use implicit differentiation: \[\frac{d}{dx}(2x^2 + 3y^2) = \frac{d}{dx}(6)\] \[4x + 6yy' = 0\] For the second curve, \(ax^2 + 4y^2 = 4\), use implicit differentiation: \[\frac{d}{dx}(ax^2 + 4y^2) = \frac{d}{dx}(4)\] \[2ax + 8yy' = 0\]

Solve for the slopes

Solve the equations from Step 1 for the slopes, which are represented by y': From the first equation: \[yy' = -\frac{2}{3}x\] \[y' = -\frac{2}{3}\frac{x}{y}\] From the second equation: \[yy' = -\frac{ax}{4}\] \[y' = -\frac{a}{4}\frac{x}{y}\]

Orthogonality condition

Orthogonality of two curves means the product of their slopes (derivatives) is -1. Substitute y' from both equations and set their product equal to -1: \[\left(-\frac{2}{3}\frac{x}{y}\right)\left(-\frac{a}{4}\frac{x}{y}\right) = -1\] It's important to note that x and y are both nonzero, as this condition holds at the intersection point of both curves.

Solve for 'a'

Solve the equation from Step 3 to find the value of 'a': \[\frac{ax^2}{4y^2} = \frac{2x^2}{3y^2}\] \[a = \frac{8x^2}{3y^2}\] To determine a, we will substitute for x^2 and y^2 from either one of the original curve equations. From the first curve: \[x^2 = \frac{6-3y^2}{2}\] Substitute this in the expression for 'a': \[a = \frac{8(6-3y^2)}{3y^2(2)}\] By simplifying, \[a = \frac{4(6-3y^2)}{y^2}\] Since we know that the two curves intersect at one point, there must be a common point (x, y) that satisfies both equations. Solving for y, we have: \[\frac{4(3y^2 - 6)}{y^2} = \frac{4(6 - 3y^2)}{y^2}\] \[3y^2 - 6 = 6 - 3y^2\] \[6y^2 = 12\] \[y^2 = 2\] Substitute this value of y^2 back into the expression for 'a': \[a = \frac{4(6-3(2))}{2}\] \[a = \frac{4(6-6)}{2}\] \[a = \boxed{\text{(b) } 1}\] So, the value of 'a' is 1 for the curves to intersect orthogonally.

Key concepts

Implicit Differentiation

Implicit differentiation is a technique used in calculus to find the derivative of a function y that is not expressed explicitly as a function of x. Instead, y is part of an equation that defines a relationship between x and y. In the case of curves defined by equations such as the ones given in the exercise (\(2x^2 + 3y^2 = 6\) and \(ax^2 + 4y^2 = 4\) ), the derivatives with respect to x of y cannot be directly computed since y is not isolated.

To find the slope of such curves, we differentiate both sides of the equation with respect to x and solve for the derivative of y, denoted as y'. In the exercise, the first curve's differentiation yields the equation \[4x + 6yy' = 0\], and the second one gives \[2ax + 8yy' = 0\]. By implicitly differentiating, we manage to express the slope of the curve (y') in terms of x and y, which will later be used to examine the orthogonality condition.

Slope of a Curve

The slope of a curve at any given point is the rate at which y changes with respect to x, which can also be understood as the steepness or incline of the curve at that point. It is represented by the derivative y' of the function at that point. In the context of the given exercise, after differentiating the equations implicitly, we derive two equations displaying the slopes of their respective curves: \[y' = -\frac{2}{3}\frac{x}{y}\] for the first curve, and \[y' = -\frac{a}{4}\frac{x}{y}\] for the second curve.

The idea is that the slope of a curve tells us how fast y is changing with respect to x. Finding the slope is crucial when studying how curves intersect because it helps us understand their behavior at the points of intersection. A positive slope indicates that the curve is increasing, while a negative slope indicates that it is decreasing. Zero slope suggests that the curve is level at that particular point.

Orthogonal Curves Condition

The condition for curves to intersect orthogonally (at a right angle) is quite simple yet essential. Two curves are orthogonal at their intersection points if the product of their slopes at these points is equal to -1. In mathematical terms, this is expressed as \[y'_1 \cdot y'_2 = -1\], where y'_1 and y'_2 are the slopes of the first and second curve, respectively, at the intersection point.

In our exercise, we're given that the curves intersect orthogonally. So after determining the slopes using implicit differentiation, we set their product equal to -1 and solve for the variable of interest, in this case, 'a'. This gives us the equation \[\left(-\frac{2}{3}\frac{x}{y}\right)\left(-\frac{a}{4}\frac{x}{y}\right) = -1\]. Simplifying this equation provides the value of 'a' that satisfies the orthogonality condition. Understanding the relationship between the slopes of two curves and how they can be used to determine orthogonality is a powerful tool in solving problems involving intersecting curves.