Problem 22
Question
Show that a portion of a tangent to a parabola intercepted between directrix and the curve subtends a right angle at the focus.
Step-by-Step Solution
Verified Answer
Question: Prove that a tangent to a parabola cuts a right angle at the focus when the portion of the tangent is intercepted between the directrix and the curve.
Answer: To prove this statement, follow these steps:
1. Consider a parabola with a vertex at the origin, the focus at point F(h,0), and the directrix at x=-h with the parabolic equation y^2 = 4hx.
2. Locate the tangent and intersecting points on the parabola (point A) and directrix (point B).
3. Differentiate the parabolic equation to find the slope of the tangent at point A.
4. Find the equation of line AB using the slope and point A.
5. Draw the triangle AFB and connect all vertices.
6. Verify the perpendicularity of lines AF and BF by checking if their slopes multiply together to equal -1.
If the product of slopes is -1, angle AFB is a right angle, therefore the tangent intercepted between the directrix and the curve subtends a right angle at the focus.
1Step 1: Understand the Elements of the Parabola
Let's consider a parabola with a vertex at the origin, the focus at point F(h,0), and the directrix at x=-h. The equation of this parabola can be written as y^2 = 4hx.
2Step 2: Locate the Tangent and Intersecting Points
Let A be a point on the parabola, and B be a point on the directrix such that AB is tangent to the parabola at A. The tangent line at point A can be obtained by differentiating y^2 = 4hx w.r.t x and substituting the coordinates of point A.
3Step 3: Differentiate the Parabola Equation and Find the Slope
Differentiate y^2 = 4hx with respect to x to get: 2y(dy/dx) = 4h. Now find the value of dy/dx at the point A. This will give us the slope of the tangent line at A.
4Step 4: Find the Equation of Line AB
By using the slope of tangent line and the coordinates of point A, we can find the equation of line AB using point-slope form.
5Step 5: Draw the triangle AFB
Draw a line segment BF and join all vertices to get a triangle AFB. We need to show that angle AFB = 90°.
6Step 6: Verify the Perpendicularity of Lines AF and BF
Two lines are perpendicular when their slopes multiply together to equal -1. Compute the slopes of the lines AF and BF and check if their product is equal to -1.
If the product of slopes is -1, we can conclude that angle AFB is a right angle, proving that a portion of the tangent intercepted between the directrix and the curve subtends a right angle at the focus.
Key Concepts
Analytical GeometryParabola Equation DifferentiationSlope of Tangent Line
Analytical Geometry
Analytical geometry, often called coordinate geometry, is the mathematical study of geometry using a coordinate system. This allows for a numerical representation of geometrical shapes and an algebraic approach to solving geometric problems. In the context of parabolas, analytical geometry provides a powerful way to describe and explore their properties using equations.
For instance, a parabola is the locus of all points that are equidistant from a fixed point, known as the focus, and a line called the directrix. The standard equation of a parabola with its vertex at the origin and the focus on the x-axis is given by \(y^2 = 4ax\), where \(a\) is the distance from the vertex to the focus. This equation can be used to determine various properties of the parabola, such as its tangent lines, axis of symmetry, and the specific shape of the curve.
For instance, a parabola is the locus of all points that are equidistant from a fixed point, known as the focus, and a line called the directrix. The standard equation of a parabola with its vertex at the origin and the focus on the x-axis is given by \(y^2 = 4ax\), where \(a\) is the distance from the vertex to the focus. This equation can be used to determine various properties of the parabola, such as its tangent lines, axis of symmetry, and the specific shape of the curve.
Parabola Equation Differentiation
Differentiating the equation of a parabola is key to finding the slope of its tangent lines at any given point. This process involves applying calculus to the geometric equation of the parabola to derive the relationship between the variables.
Take for example the parabola with equation \(y^2 = 4ax\). By differentiating both sides with respect to \(x\), the derivative \(2y(dy/dx) = 4a\) is obtained. This expression can be rearranged to find the gradient or slope of the tangent line by isolating \(dy/dx\), which gives \(dy/dx = \frac{2a}{y}\). The slope of the tangent line at any point on the parabola can then be calculated by substituting the y-coordinate of that point into the derivative. This differentiation is pivotal not only in finding slopes but also when analyzing motion along the parabola's path or determining instantaneous rates of change.
Take for example the parabola with equation \(y^2 = 4ax\). By differentiating both sides with respect to \(x\), the derivative \(2y(dy/dx) = 4a\) is obtained. This expression can be rearranged to find the gradient or slope of the tangent line by isolating \(dy/dx\), which gives \(dy/dx = \frac{2a}{y}\). The slope of the tangent line at any point on the parabola can then be calculated by substituting the y-coordinate of that point into the derivative. This differentiation is pivotal not only in finding slopes but also when analyzing motion along the parabola's path or determining instantaneous rates of change.
Slope of Tangent Line
The slope of a tangent line to a curve at a given point is essentially the instantaneous rate of change of the curve at that point. For a parabola, once we have the derivative from differentiating the parabola's equation, we can find the slope at any particular point by substituting the coordinates of that point into the derivative formula.
To find the slope of a tangent line at a point \(A(x_1, y_1)\) on the parabola, we substitute \(y_1\) into the derivative to get \(dy/dx|_{(x_1, y_1)} = \frac{2a}{y_1}\), where the vertical bar indicates evaluation at point \(A\). This slope is an essential part of determining the equation of the tangent line itself, using the point-slope form \(y - y_1 = m(x - x_1)\), where \(m\) represents the slope. Understanding the slope is also crucial for solving geometric problems such as the one in our exercise, which involves proving that a line segment subtends a right angle with respect to a point on the curve.
To find the slope of a tangent line at a point \(A(x_1, y_1)\) on the parabola, we substitute \(y_1\) into the derivative to get \(dy/dx|_{(x_1, y_1)} = \frac{2a}{y_1}\), where the vertical bar indicates evaluation at point \(A\). This slope is an essential part of determining the equation of the tangent line itself, using the point-slope form \(y - y_1 = m(x - x_1)\), where \(m\) represents the slope. Understanding the slope is also crucial for solving geometric problems such as the one in our exercise, which involves proving that a line segment subtends a right angle with respect to a point on the curve.
Other exercises in this chapter
Problem 20
Prove that the tangent to a parabola and the perpendicular to it from its focus meet on the tangent at the vertex.
View solution Problem 21
Show that a portion of a tangent to a parabola intercepted between directrix and the curve subtends a right angle at the focus.
View solution Problem 23
The tangent to the parabola \(y^{2}=4 a x\) make angles \(\theta_{1}\) and \(\theta_{2}\) with the axIs. Show that the locus of the point of intersection such t
View solution Problem 25
Prove that the equation of the parabola whose vertex and focus on \(x\) -axis at distances \(4 a\) and \(5 a\) from the origin respectively \((a>0)\) is \(y^{2}
View solution