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Chapter 4: Problem 3

A: The ratio of length of tangent to length of normal is proportional to the ordinate of the point of tangency at the curve \(y^{2}=4 a x\). R: Length of normal and tangent to a curve \(y=f(x)\) is \(|y \sqrt{m}|\) and \(\left|\frac{y \sqrt{1+m^{2}}}{m}\right|\), where \(m=\frac{d y}{d x}\).

Short Answer

Expert verified
In conclusion, the ratio of the lengths of the tangent and normal for the curve \(y^{2}=4 a x\) is proportional to the ordinate (y-coordinate) of the point of tangency. This can be expressed as: \[\frac{|Tangent|}{|Normal|} \propto y\]

Step by step solution

01

Differentiate the given equation

First, we will find the derivative of the given equation, \(y^2 = 4ax\), concerning \(x\). This will give us the slope (\(m\)) of the tangent for any point on the curve. To differentiate, we will apply the chain rule. \(2y\frac{dy}{dx} = 4a\) Now, we can solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{2a}{y}\)
02

Find the length of the tangent

We're given that the length of the tangent to a curve is given by the formula \(\left|\frac{y \sqrt{1+m^{2}}}{m}\right|\). We already have the value of \(m\), so let's substitute it into the formula: \[|Tangent| = \left|\frac{y \sqrt{1 + \left(\frac{2a}{y}\right)^2}}{\frac{2a}{y}}\right|\] By simplifying the expression, we find: \[|Tangent| = \left|\frac{y\sqrt{y^2+4a^2}}{2a}\right|\]
03

Find the length of the normal

We're given that the length of the normal to a curve is given by the formula \(|y \sqrt{m}|\). We already have the value of \(m\), so let's substitute it into the formula: \[|Normal| = \left|y\sqrt{\frac{2a}{y}}\right|\] By simplifying the expression, we find: \[|Normal| = \left|\frac{y\sqrt{2a}}{\sqrt{y}}\right|\]
04

Find the ratio of the lengths

Now, we need to find the ratio of the length of the tangent to the length of the normal. We will divide the expression for the length of the tangent by the expression for the length of the normal: \[\frac{|Tangent|}{|Normal|} = \frac{\left|\frac{y\sqrt{y^2+4a^2}}{2a}\right|}{\left|\frac{y\sqrt{2a}}{\sqrt{y}}\right|}\] Simplifying the ratio, we find: \[\frac{|Tangent|}{|Normal|} = \frac{y\sqrt{y^2+4a^2}\cdot\sqrt{y}}{2a\cdot y\sqrt{2a}}\] This can be further simplified to: \[\frac{|Tangent|}{|Normal|} = \frac{y^2\sqrt{y^2+4a^2}}{2ay\sqrt{2a}}\] Since the problem statement indicates that this ratio is proportional to the ordinate (y-coordinate) of the point of tangency, we can conclude that: \[\frac{|Tangent|}{|Normal|} \propto y\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Calculus
Differential calculus is the branch of mathematics that focuses on rates of change and slopes of curves. It is used to determine the gradient of a tangent line to a curve at any given point. This is pivotal in understanding how a function behaves and how it can be optimized or analyzed.

For any function represented by an equation, differentiation gives us the derivative, often denoted by \( \frac{dy}{dx} \). The derivative reflects how the output of the function (generally 'y') changes with respect to changes in the input (usually 'x'). In the provided exercise, the derivative of \( y^2 = 4ax \) with respect to \( x \) was found using the chain rule, resulting in \( \frac{dy}{dx} = \frac{2a}{y} \). This gradient, \( m \), is then used in further calculations to determine the lengths of tangents and normals to the curve at a specific point.
Curve Tangency
Curve tangency refers to the point where a tangent line just touches a curve without intersecting it at that immediate vicinity. The tangent line is a straight line that represents the direction of the curve at the point of tangency. It's the best linear approximation of the curve near that point.

The slope of this tangent line at any point on the curve is equal to the derivative of the curve's equation at that point. As we saw in the exercise, after finding the slope, we used the formula for the length of the tangent with respect to this slope, which utilizes the principles of tangency to find the precise length.
Length of Tangent
The length of a tangent to a curve at a particular point is the distance from that point to an imaginary point where the tangent line ceases to exist or intersects the curve again, which conceptually is at infinity. In mathematical terms, we can measure this length in a finite way by considering geometrical shapes like triangles formed by the tangent line, the radius from the curve, and other reference lines.

In the solution provided, the length of tangent was calculated using the formula \( |Tangent| = \left|\frac{y \sqrt{1+m^{2}}}{m}\right| \) after determining the slope \( m = \frac{dy}{dx} \). Understanding how to compute the length of the tangent is fundamental in geometry and calculus since it encapsulates the relationship between a curve and its tangent at a specific point.
Length of Normal
The length of the normal is the distance from the point of tangency to the curve perpendicularly until it reaches either the x-axis or the extension of the radius that passed through the point of tangency. The normal is orthogonal to the tangent and provides a direct geometric and algebraic relationship to the curve's curvature at a particular point.

In the given problem's context, we used the length of normal formula \( |Normal| = |y \sqrt{m}| \) to calculate the normal length. The normal plays an essential role in optimization problems and in constructing geometric shapes related to the curve. It鈥檚 especially interesting in the context of the original exercise, as the tangent-to-normal ratio reflects properties of the curve related to the specific point of tangency.

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Most popular questions from this chapter

A curve \(C\) has the property that if the tangent drawn at any point \(P\) on \(C\) meets the co-ordinate axes at \(A\) and \(B\), then \(P\) is the mid-point of \(A B\). The curve passes through the point \((1,1)\). The equation of the curve is (a) \(x y=1\) (b) \(y^{2}=2 x-1\) (c) \(x^{2}=2 y-1\) (d) None of these

For the curve represented paramatrically by the equations, \(x=2 \ell n \cot t+1 \& y=\tan t+\cot t\) (a) tangent at \(t=\pi / 4\) is parallel to \(x\)-axis (b) normal at \(t=\pi / 4\) is parallel to \(y\)-axis (c) tangent at \(t=\pi / 4\) is parallel to the line \(y=x\) (d) tangent and normal intersect at the point \((2,1)\)

If the curve \(\left(\frac{x}{a}\right)^{n}+\left(\frac{y}{b}\right)^{n}=2\) touches the straight line \(\frac{x}{a}+\frac{y}{b}=2\), then find the value of ' \(n\) ' (a) 2 (b) 3 (c) 4 (d) any real number

Show that if \(P(x)\) is a polynomial of odd degree greater than 1 , then through any point \(P\) in the plane, there will be atleast one tangent line to the curve \(y=P(x)\). Is this true if \(P(x)\) is a curve of even degree? Solution: If \(y=P(x)\), where \(P(x)\) is a polynomial of odd degree \(d>1\), then through any point \(P(a, b)\) there will be atleast one tangent to the curve if and only if \((y-b)=P^{\prime}(x)(x-a)\) or \(\\{P(x)-b\\}=P^{\prime}(x)\\{(x-a)\\}\) has a real solution But \(x P^{\prime}(x)-P(x)-a P^{\prime}(x)+b=0\) has a degree ' \(d\) with leading coefficient \((d-1)\) times the leadingcoefficient of \(P(x)\) and by intermediate value theorem it has root (real) i.e., There is a real number \(x_{0}\) for which tangent to \(y=P(x)\) at \(\left\\{x_{0}, P\left(x_{0}\right)\right.\) passes through \(P(a, b) .\) For even degree it may not be true (Consider \(y=x^{2}\) ).

Consider the curve represented parametrically by the equation \(x=t^{3}-4 t^{2}-3 t\) and \(y=2 t^{2}+3 t-5\), where \(t \in \mathbb{R} .\) If \(H\) denotes the number of point on the curve where the tangent is horizontal and \(V\) the number of point where the tangent is vertical, then (a) \(H=2\) and \(V=1\) (b) \(H=1\) and \(V=2\) (c) \(H=2\) and \(V=2\) (d) \(H=1\) and \(V=1\)
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