Find equations of both lines through the point (2, ?3) that are tangent to the parabola y = x^2 + x. y_1=(smaller slope quation) y_2=(larger slope equation)

Falak Kinney

Falak Kinney

Answered question

2021-05-17

Find equations of both lines through the point (2, ?3) that are tangent to the parabola y=x2+x.
y1=(smaller slope quation)
y2=(larger slope equation)

Answer & Explanation

oppturf

oppturf

Skilled2021-05-18Added 94 answers

The answer to your question:

 

image
image

Jeffrey Jordon

Jeffrey Jordon

Expert2021-09-09Added 2605 answers

Differentiate y=x2+x with respect to x

dydx=ddx(x2+x)=ddx(x2)+ddx(x)

By using the Powe Rule ddx(xn)=nxn1

dydx=2x21+x11

=2x+1

Since y=x2+x, consider a point on the parabola as (a,a2+a).

The slope of the line at (a,a2+a) is m=dydx=2a+1

Use the slope-intercept form y=mx+cy=(2a+1)x+C

If this line is a tangent to the give parabola, then it passes through (a,a2+a), so

a2+a=(2a+1)a+c

a2+a=2a2+a+c

c=a2

Therefore the equation of the tangent line is y=(2a+1)xa2

Since the required tangent passes through the point (2,3)

y=(2a+1)xa2

3=(2a+1)2a2

a24a5=0

a=1 or a=5

From equation (1):

if  a=1, then

y=(2a+1)xa2

y=(2+1)x(1)2

y1=x1

if a=5, then

y=(2a+1)xa2

y=(10+1)x52

y2=11x25

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?