Question

Show that the graphs of \(2 x^{2}+y^{2}=6\) and \(y^{2}=4 x\) intersect at right angles.

Short answer

The curves intersect at points (1, 2) and (1, -2), forming right angles.

Step-by-step solution

Find Points of Intersection

To find where the curves intersect, solve the system of equations given by \(2x^2 + y^2 = 6\) and \(y^2 = 4x\). Substitute \(y^2 = 4x\) into the first equation: \(2x^2 + 4x = 6\). Simplify it to \(x^2 + 2x - 3 = 0\). Factor the quadratic: \((x + 3)(x - 1) = 0\). Thus, \(x = -3\) or \(x = 1\). Substitute back into \(y^2 = 4x\) to get the corresponding \(y\) values. For \(x = 1\), \(y^2 = 4 \times 1 = 4\), so \(y = \pm2\). The point is \((1, 2)\) and \((1, -2)\). For \(x = -3\), \(y^2 = 4 \times (-3)\) which is not valid since \(y^2\) cannot be negative. Hence, the points of intersection are \((1, 2)\) and \((1, -2)\).

Determine Tangents at Intersection Points

To determine if the curves intersect at right angles, we need the slopes of the tangents at the intersection points. Find the derivatives. For \(2x^2 + y^2 = 6\), implicitly differentiate to get \(4x + 2y\frac{dy}{dx} = 0\). Thus, \(\frac{dy}{dx} = -\frac{2x}{y}\). For \(y^2 = 4x\), differentiate to get \(2y\frac{dy}{dx} = 4\). Thus, \(\frac{dy}{dx} = \frac{2}{y}\).

Evaluate Slopes at Intersection Points

At point \((1, 2)\), substitute into the slopes. For \(2x^2 + y^2 = 6\), \(\frac{dy}{dx} = -\frac{2 \cdot 1}{2} = -1\). For \(y^2 = 4x\), \(\frac{dy}{dx} = \frac{2}{2} = 1\). At point \((1, -2)\), for \(2x^2 + y^2 = 6\), \(\frac{dy}{dx} = -\frac{2 \cdot 1}{-2} = 1\). For \(y^2 = 4x\), \(\frac{dy}{dx} = \frac{2}{-2} = -1\).

Confirm Right Angle Intersection

Two lines are perpendicular (intersect at right angles) if the product of their slopes is -1. For both intersection points \((1, 2)\) and \((1, -2)\), the product of the slopes \(-1 \times 1 = -1\) and \(1 \times -1 = -1\) confirms the curves intersect at right angles.

Key concepts

Tangent Lines

Tangent lines are straight lines that lightly touch a curve at a particular point, without crossing over it. They provide an instantaneous direction, much like how a velocity vector shows direction and speed for a moving object. When you find the tangent line's slope at a specific point on a curve, you're actually examining the curve's slope at that point.

In the context of our exercise, once we understand where the two curves intersect, we must determine the slope of each curve's tangent at those points to confirm they intersect at right angles. This requires us to first understand and then calculate the derivatives of the functions involved, which leads us directly into the concept of implicit differentiation.

Implicit Differentiation

Implicit differentiation is a technique used when you have an equation where the dependent and independent variables are intertwined. This type of differentiation is essential when you cannot easily solve for one variable in terms of the other.

For the curves given in the exercise, i.e., the ellipse-like equation and the parabola-like equation, both involve terms with both x and y. Specifically, the equation, \(2x^2 + y^2 = 6\), does not separate as y equals some function of x directly.

To find the derivative of y with respect to x (dy/dx), you perform implicit differentiation by treating y as a function of x. When differentiating terms involving y, remember to multiply by dy/dx per the chain rule. This allows us to find the slope for the tangent line of the curve at any given point, which is crucial in demonstrating whether the curves are perpendicular at their points of intersection.

Quadratic Equations

Quadratic equations are fundamental mathematical expressions that are always in the form ax² + bx + c = 0, where a, b, and c are constants. Solving quadratic equations can tell you about the x-intercepts of a graph or even points of intersection between different curves.

In our exercise, after substituting one equation into another, we ended up with a quadratic equation: \( x^2 + 2x - 3 = 0 \). Solving this helps in finding potential x-values where the curves intercept. The process is to factorize, use the quadratic formula, or complete the square to find the roots of the equation.

For this case, the equation factors nicely into (x + 3)(x - 1) = 0, leading to x-values of -3 and 1. These values are essential for determining the exact points where the intersection occurs, allowing us to further test if their slopes are perpendicular.

Perpendicular Slopes

Perpendicular slopes represent lines that intersect at a 90-degree angle. A quick test for perpendicularity, when dealing with slopes, is to multiply them. For two slopes to be perpendicular, their product must equal -1.

In our context, after finding the slopes of the tangents at the points of intersection using implicit differentiation, we obtained values of 1 and -1 at each intersection point.

By checking

• if \(-1 \times 1 = -1\)
• if \(1 \times -1 = -1\)

for both calculated pairs, we verified that the curves indeed intersect at right angles. This neat trick simplifies checking the perpendicular nature of intersecting curves, essentially confirming orthogonality quickly and elegantly.