Question

find the equation of the tangent and the normal to the curve x²+y²=5 where the tangent is parallel to line 6x+3y-5=0​

Answer 1

Final Answer:

The required equations of the tangents to the curve $x^2+y^2=5$ where the tangent is parallel to line $6x+3y-5=0$ are $2x-y+5=0,2x-y-5=0$ and the equation of the normal to the curve $x^2+y^2=5$ is $x+2y=0$.

Given:

The equation of the curve is $x^2+y^2=5$ and the other equation of the line is $6x+3y-5=0$ .

To Find:

The equation of the tangent and the equation of the normal to the curve $x^2+y^2=5$ where the tangent is parallel to line $6x+3y-5=0$​

Explanation:

The tangent to any curve is the straight line drawn to the curve from outside, such that the straight line touches the curve at a single point of contact.

The normal to the curve is the perpendicular line which is at a right angle to the tangent of the curve at the point of its contact with the curve.

The slope-intercept form of the equation of a straight line is $y=mx+c$.

Step 1 of 6

Differentiate the equation of the curve is $x^2+y^2=5$  with respect to x.

$\frac{d}{dx} (x^2)+\frac{d}{dx} (y^2)=\frac{d}{dx} (5)\\2x+2y\frac{dy}{dx}=0\\2y\frac{dy}{dy}=-2x\\\frac{dy}{dx}=-\frac{2x}{2y}\\\frac{dy}{dx}=-\frac{x}{y}$

Step 2 of 6

Again, rewrite the equation of the curve in the following way.

$x^2+y^2=5\\y^2=5-x^2\\y=\sqrt{5-x^2$

Differentiate this equation with respect to x.

$\frac{d}{dx} y=\frac{d}{dx}\sqrt{(5-x^2)}\\\frac{dy}{dx}=\frac{(0-2x)}{2\sqrt{(5-x^2)}}\\\frac{dy}{dx}=\frac{x}{\sqrt{(5-x^2)}}$

Step 3 of 6

The slope of the line $6x+3y-5=0$ is written in the following way.

$6x+3y-5=0\\3y=-6x+5\\y=-2x+\frac{5}{3}$

The slope of the line $6x+3y-5=0$ is -2. Now, the line parallel to the line $6x+3y-5=0$ will have the same slope which is equal to -2.

Step 4 of 6

Let us assume that $(x_1 ,y_2)$ be the point of contact of the tangent. Thus it satisfies both the differentiation equations and we get the following.

$\frac{dy}{dx}=-\frac{x_1}{y_1}\\\frac{dy}{dx}=\frac{x_1}{\sqrt{(5-x_1^2)}}$

By the definition of the slope, we get the following equations.

The first equation is

$2=-\frac{x_1}{y_1}\\-x_1=2y_1\\x_1+2y_1=0$

The second equation is

$\frac{x_1}{\sqrt{(5-x_1^2)}}=2\\\frac{{x_1}^2}{(5-x_1^2)}=4\\x_1^2=20-4x_1^2\\5x_1^2=20\\x_1^2=4\\x_1=\pm 2$

Step 5 of 6

Solving the above equations, we get the following coordinates of $(x_1 ,y_2)$.

$x_1=2,y_1=-1\\x_1=-2, y_1=1$

Thus, the equations of the tangents to the curve $x^2+y^2=5$ where the tangent is parallel to line $6x+3y-5=0$ are given by the following.

The first equation of the tangent is

$y-1=2(x+2)\\y-1=2x+4\\2x-y+5=0$

The second equation of the tangent is

$y+1=2(x-2)\\y+1=2x-4\\2x-y-5=0$

Step 6 of 6

By the definition, the curve is $x^2+y^2=5$ represents a circle with its center at $(0.0)$. The normal to its tangent will, obviously, cross through the points of contact of tangents and the center of the circle.

So, the normal passing through the center $(0.0)$ and $(x_1=2,y_1=-1)$ is

$y-0=\frac{-1-0}{2-0} (x-0)\\y=-\frac{1}{2} x\\2y=-x\\x+2y=0$

Again, the normal passing through the center $(0.0)$ and $(x_1=-2, y_1=1)$

$y-0=\frac{1-0}{-2-0} (x-0)\\y=-\frac{1}{2} x\\2y=-x\\x+2y=0$

Therefore, the required equations of the tangents to the curve $x^2+y^2=5$ where the tangent is parallel to line $6x+3y-5=0$ are $2x-y+5=0,2x-y-5=0$ and the required equation of the normal to the curve $x^2+y^2=5$  is $x+2y=0$.

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