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Chapter 13: Problem 1101

Order and degree of differential equation of all tangent lines to the parabola \(\mathrm{y}^{2}=4 \mathrm{ax}\) is (A) 2,2 (B) 3,1 (C) 1,2 (D) 4,1

Short Answer

Expert verified
The order and degree of the differential equation of all tangent lines to the parabola \(y^2 = 4ax\) is 1 and 1, respectively. However, none of the provided options (A, B, C, D) match the correct answer.

Step by step solution

01

Find the general equation of the tangent line

First, we need to find the slope of the tangent line to the parabola \(y^2 = 4ax\). To do this, we'll differentiate the equation implicitly with respect to x: \[\frac{d}{dx}(y^2) = \frac{d}{dx}(4ax)\] Applying the chain rule, we get: \[2y \frac{dy}{dx} = 4a\] Now, we can solve for the slope, \( \frac{dy}{dx}\): \[\frac{dy}{dx} = \frac{4a}{2y}\] The slope of the tangent line is \(\frac{4a}{2y}\), and we know that the line touches the parabola at a point \((x, y)\). Using the point-slope form of a linear equation, we can write the general equation of the tangent line: \[y - y_1=\frac{4a}{2y_1}(x-x_1)\], where \((x_1, y_1)\) is the point on the parabola where the tangent line touches.
02

Eliminate the arbitrary constants

Now, we need to eliminate the arbitrary constants, that is, x1 and y1, from the tangent line equation. Since \((x_1, y_1)\) is on the parabola, we have \[y_1^2 = 4ax_1\] Now express x1 in terms of y1: \[x_1 = \frac{y_1^2}{4a}\] Substitute this back into the tangent line equation: \[y - y_1 = \frac{4a}{2y_1}(x - \frac{y_1^2}{4a})\]
03

Find the differential equation and its order and degree

Simplify the equation and express in terms of derivatives: \[y - y_1 = \frac{dy}{dx}(x - \frac{y^2}{4a})\] \[y - y_1 = (x\frac{dy}{dx} - \frac{y^2}{4}\frac{dy}{dx})\] Now multiply both sides by 2y and rearrange: \[0 = 2y\frac{dy}{dx}x - y^2\frac{dy}{dx} - 2yy_1\] Notice that this is a first-order differential equation because there is only one derivative, and the highest power of the derivative is 1, so the degree is also 1. Comparing to the given options, we find that the order and degree of the differential equation are 1 and 1, respectively. Therefore, none of the provided options (A, B, C, D) match the correct answer.

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Most popular questions from this chapter

If \(y=\left[x+\sqrt{ \left.\left(1+x^{2}\right)\right]^{n}}\right.\), then \(\left(1+x^{2}\right) \cdot\left(d^{2} y / d x^{2}\right)+x \cdot(d y / d x)\) - (A) - (B) \(2 \mathrm{x}^{2} \mathrm{y}\) (C) \(\mathrm{n}^{2} \mathrm{y}\) (D) \(-\mathrm{n}^{2} \mathrm{y}\)

The differential equation of all circles passing through the origin and having their centers on the y-axis is: OR The differential equation for the family of curves \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{ay}=0\), where a is an arbitrary constant is: (A) \(\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right) \mathrm{y}^{1}=2 \mathrm{xy}\) (B) \(2\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right) \mathrm{y}^{1}=\mathrm{xy}\) (C) \(2\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right) \mathrm{y}^{1}=\mathrm{xy}\) (D) \(\left(x^{2}+y^{2}\right) y^{1}=2 x y\)

The solution of \(\mathrm{x}^{2} \mathrm{y}-\mathrm{x}^{3}(\mathrm{dy} / \mathrm{dx})=\mathrm{y}^{4} \cos \mathrm{x} ; \mathrm{y}(0)=1\) is: (A) \(x^{3}=y^{3} \sin x\) (B) \(x^{3}=3 y^{3} \sin x\) (C) \(y^{3}=3 x^{3} \sin x\) (D) none the these

The Integrating factor of the differential equation \(\left(1-\mathrm{y}^{2}\right)(\mathrm{dy} / \mathrm{dx})-\mathrm{yx}=1\) is: (A) \(\left[1 / \sqrt{ \left.\left(1-\mathrm{y}^{2}\right)\right]}\right.\) (B) \(\sqrt{\left(1-\mathrm{y}^{2}\right)}\) (C) \(\left[1 /\left(1-\mathrm{y}^{2}\right)\right]\) (D) \(1-\mathrm{y}^{2}\)

The particular solution of \(\left(1+\mathrm{y}^{2}\right) \mathrm{dx}+\left(\mathrm{x}-\mathrm{e}^{-(\tan )-1 \mathrm{y}}\right) \mathrm{dy}=0\) with initial condition \(\mathrm{y}(0)=0\) is: (A) \(x \cdot e^{(\tan )-1 x}=\cot ^{-1} x\) (B) \(x \cdot e^{(\tan )-1 y}=\tan ^{-1} y\) (C) \(x \cdot e^{(\tan )-1 y}=\cot ^{-1} y\) (D) \(x \cdot e^{(\cot )-1 y}=\tan ^{-1} y\)
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