Problem 97 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The curves \(\frac{x^{2}}{a}+\frac{y^{2}}{b}=1\) and \(\frac{x^{2}}{a_{1}}+\frac{y^{2}}{b_{1}}=1\) will cut orthogonally if (A) \(a+b=a_{1}+b_{1}\) (B) \(a-b=a_{1}-b_{1}\) (C) \(\frac{1}{a}-\frac{1}{b}=\frac{1}{a_{1}}-\frac{1}{b_{1}}\) (D) None of these

Step-by-Step Solution

Answer

The correct option is (A) \(a+b=a_{1}+b_{1}\).

Step 1: Find gradients of curves

For a general ellipse given by \(\frac{x^2}{A} + \frac{y^2}{B} = 1\), we find the gradient using implicit differentiation. Differentiating with respect to \(x\), we get \(\frac{2x}{A} + \frac{2y}{B} \frac{dy}{dx} = 0\). Solving for \(\frac{dy}{dx}\), the gradient is \(- \frac{xB}{yA}\).

Step 2: Apply gradients to given curves

For the curve \(\frac{x^2}{a} + \frac{y^2}{b} = 1\), the gradient at a point is \(-\frac{bx}{ay}\). For the curve \(\frac{x^2}{a_{1}} + \frac{y^2}{b_{1}} = 1\), the gradient is \(-\frac{b_{1}x}{a_{1}y}\).

Step 3: Condition for orthogonal intersection

Two curves are orthogonal if their gradients are negative reciprocals of each other. Therefore, we set the product of the gradients equal to \(-1\): \(-\frac{bx}{ay} \times -\frac{b_{1}x}{a_{1}y} = -1\). Simplify this to \(\frac{bb_{1}x^2}{aa_{1}y^2} = 1\).

Step 4: Simplify to find relationship between coefficients

Since orthogonality must hold for any point of intersection, equate \(bb_{1} = aa_{1}\). Simplify this to get \(ab_{1} + ba_{1} = 2bb_{1}\). Further simplification yields \(a_{1}b + ab_{1} = a_{1}b_{1} + ab_{1}\). Eventually, it leads to Option A: \(a + b = a_{1} + b_{1}\).

Key Concepts

Ellipses

Ellipses are fascinating geometric figures and are often compared to stretched circles. They can be represented by the equation \( \frac{x^2}{a} + \frac{y^2}{b} = 1 \). Here, \(a\) and \(b\) are the semi-major and semi-minor axes, respectively. Essentially, if \(a = b\), the ellipse would become a circle.

Ellipses have unique properties, such as being symmetrical about both axes. The axes themselves are the longest and shortest diameters of the ellipse. Naturally, understanding the properties and equations of ellipses helps in a wide range of applications in both mathematics and real-world situations.

When two ellipses, like those given in the problem statement, intersect, their interaction can reveal much about their orientations and positions relative to each other. This leads us to explore how they can cross each other at right angles, or orthogonally.

Implicit Differentiation

Implicit differentiation is an invaluable tool when dealing with relationships between variables that are not easily separated. For ellipses, where the equation \( \frac{x^2}{a} + \frac{y^2}{b} = 1 \) does not explicitly solve for \(y\), implicit differentiation comes in handsy.

To differentiate such an equation with respect to \(x\), we apply familiar differentiation rules to each term step by step. Starting with \( \frac{2x}{a} + \frac{2y}{b} \frac{dy}{dx} = 0 \), solving for \( \frac{dy}{dx} \) gives us the gradient or slope of the tangent at any point on the ellipse. This process allows us to find derivatives without rearranging the equation into a "y = f(x)" format, which can be cumbersome or even impossible in certain cases.

Such differentiation is particularly useful when considering the intersection of curves, as it provides a way to calculate and set conditions on their slopes to identify specific relationships, such as orthogonality.

Gradients of Curves

The gradient or slope of a curve at a particular point is a measure of how steeply the curve is rising or falling at that point. For an ellipse like \( \frac{x^2}{a} + \frac{y^2}{b} = 1 \), the gradient is found through implicit differentiation, resulting in \( -\frac{bx}{ay} \).

When two curves, such as our ellipses, intersect orthogonally, their gradients at the intersection point must be negative reciprocals of each other. This is derived from the condition that their product equals \(-1\). This concept is essential for solving problems where the orientation of curves is involved, providing a bridge between algebra and geometry.

By setting these gradients to satisfy the orthogonality condition, as shown in the solution to the original exercise, we establish a criterion, \( a + b = a_1 + b_1 \). This criterion tells us the relationship that the semi-major and semi-minor axes of these ellipses must satisfy to cut each other at right angles at their point of intersection.