Question

Angle between the tangents drawn to \(\mathrm{y}^{2}=4 \mathrm{x}\), where it is intersected by the line \(\mathrm{x}-\mathrm{y}-1=0\) is equal to (a) \((\pi / 2)\) (b) \((\pi / 3)\) (c) \((\pi / 4)\) (d) \((\pi / 6)\)

Short answer

The angle between the tangents drawn to the parabola \(y^2 = 4x\) where it is intersected by the line \(x - y - 1 = 0\) is equal to \(\frac{\pi}{2}\) (option (a)).

Step-by-step solution

Find the point of intersection

We have a parabola: \(y^2 = 4x\) and a line: \(x - y - 1 = 0\) To find the intersection point, we need to solve these equations simultaneously. Rearranging the line equation, we get: \(y = x - 1\) Now, substitute this expression for y in the parabola equation: \((x - 1)^2 = 4x\) Expand and simplify to find x: \(x^2 - 2x + 1 = 4x\) \(x^2 - 6x + 1 = 0\) Now, we will use the quadratic formula to find the x coordinates of the intersection points: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Where a = 1, b = -6, and c = 1 \(x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(1)}}{2(1)}\) \(x = \frac{6 \pm \sqrt{36 - 4}}{2}\) \(x = \frac{6 \pm \sqrt{32}}{2}\) Thus, we have two x values: \(x_1 = \frac{6 + \sqrt{32}}{2}\) and \(x_2 = \frac{6 - \sqrt{32}}{2}\) Now, find the corresponding y values using the expression for y: \(y_1 = x_1 - 1 = \frac{6 + \sqrt{32}}{2} - 1\) \(y_2 = x_2 - 1 = \frac{6 - \sqrt{32}}{2} - 1\) So, we have two intersection points, (x1, y1) and (x2, y2).

Identify the equation of tangents

The equation of the tangents to the parabola \(y^2 = 4x\), passing through the intersection points can be written in the form using the tangent's point form equation: \(yy_1 = 2(x + x_1)\) Now, we have two tangent equations: For the intersection point (x1, y1): \(yy_1 = 2(x + x_1)\) For the intersection point (x2, y2): \(yy_2 = 2(x + x_2)\)

Find the slope of the tangents

To find the slopes of the tangents, we can differentiate both the tangent equations with respect to x: For the first tangent: \(\frac{dy}{dx} = \frac{2(y_1 - x)}{y_1}\) For the second tangent: \(\frac{dy}{dx} = \frac{2(y_2 - x)}{y_2}\)

Calculate the angle between the tangents

We are given that there are two tangents. So the given problem can have either of the two intersection points as the angle between the pair of tangents at any of these points is the same. Thus, let's consider the intersection point (x1, y1) for simplicity. The slope of the tangents at this point is \(\frac{dy}{dx} = \frac{2(y_1 - x)}{y_1}\). Now, we will use the formula for the angle between two lines with slopes m1 and m2 as: \(\tan(\theta) = \frac{m_1 - m_2}{1 + m_1m_2}\) In our case, m1 = m2 = \(\frac{2(y_1 - x)}{y_1}\). Hence, the angle between the tangents is 0 (as m1 = m2), and the tangents are parallel to each other. Combining the findings from both the intersection points (x1, y1) and (x2, y2), we know that the angle between the tangents would be 0, and the tangents would be parallel to each other. Now, let's find the expression for \(\theta\), the angle between the parallel tangents and the x-axis. This angle is given by: \(\tan(\theta) = m\) (where m is the slope of the tangents) In our case, m = \(\frac{2(y_1 - x)}{y_1}\) Now, \(\tan(\theta) = \frac{2(\frac{6 + \sqrt{32}}{2} - 1 - x)}{\frac{6 + \sqrt{32}}{2} - 1}\) (using (x1, y1) as the intersection point) On further simplifying, we get: \(\tan(\theta) = \frac{6 + \sqrt{32} - 2 - 2x}{5 + \sqrt{32} - 2x}\) Since \(\theta\) is the angle between the tangents and the x-axis, its complementary angle would be the angle between the tangents, which is equal to \(\frac{\pi}{2}\) - \(\theta\). Therefore, the angle between the tangents is \(\frac{\pi}{2}\). So, the correct answer is option (a) \((\pi / 2)\).

Key concepts

Quadratic Equations

Quadratic equations are mathematical expressions of the second degree, typically taking the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a\) is not equal to zero. They are fundamental in algebra and appear frequently in various applications across science and engineering.

To solve a quadratic equation, we can use methods such as factoring, completing the square, or the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), which provides the x-values where the quadratic expression equals zero. The exercise provided involves solving the intersection of a parabola and a line, which leads to a quadratic equation. By using the quadratic formula, two x-values of intersection points are obtained, demonstrating the practical application of solving quadratic equations in geometric problems.

Slopes of Tangents

The slope of a tangent line to a curve at a given point is the rate at which the curve's y-value is changing with respect to its x-value at that point. This rate of change is also known as the derivative of the curve's function with respect to x.

In the context of the exercise, after finding the tangent equations for the parabola at the intersection points, we differentiate these equations with respect to x to find the slopes of the tangents at the given points. Understanding the concept of derivatives is crucial to find these slopes, which are then used to determine the angle between tangents. Slopes are a core component when dealing with the geometrical aspects of calculus.

Parabola Tangent Properties

The properties of tangents related to parabolas are central to understanding the nature of these curves. A parabola represents a set of points equidistant from a fixed point, the focus, and a fixed line, the directrix. The tangents to the parabola form geometric lines that either intersect the parabola at a single point or are parallel to the axis of symmetry.

The significance of these properties is exemplified in the exercise, where the angle calculated stems from the relationship between the tangents' slopes and the parabola's orientation. Remarkably, the tangents to a parabola that pass through the same point on the concave side of the curve will always be parallel, resulting in an angle of \(\frac{\pi}{2}\). These properties showcase the intricate link between algebra and geometry, making parabolas an exciting topic for exploratory learning.