Equation of tangent to parabola

**goguy123** · January 16th 2010, 07:51 AM

Got a Prelim on monday

, please help, thanks

Q: Find the equation of the tangent to the parabola with equation
y=6+x-x(squared) at the point (1,6 ) .

If possible please explain to me how to do this question thanks .

**e^(i*pi)** · January 16th 2010, 07:58 AM

Originally Posted by goguy123

Got a Prelim on monday

, please help, thanks

Q: Find the equation of the tangent to the parabola with equation
y=6+x-x(squared) at the point (1,6 ) .

If possible please explain to me how to do this question thanks .

If you've done (or supposed to have done) calculus take the derivative and find $f'(1)$

In this case $f'(1) = -1$

Now you know the gradient use the straight line equation $y=mx+c$ to find $c$ using the ordered pair $(1,6)$

In this case $c = 7$

So the equation is $y = -x + 7$ or $y= 7-x$

**VonNemo19** · January 16th 2010, 07:59 AM

Originally Posted by goguy123

Got a Prelim on monday

, please help, thanks

Q: Find the equation of the tangent to the parabola with equation
y=6+x-x(squared) at the point (1,6 ) .

If possible please explain to me how to do this question thanks .

The slope of the line tangent to $f(x)$ at the point $(1.6)$ is $m=f'(1)$ .

So, recall that the equation to a line is $(y_2-y_1)=m(x_2-x)$

**goguy123** · January 16th 2010, 08:12 AM

Thanks a lot

for the help.

**mr fantastic** · January 16th 2010, 06:52 PM

Originally Posted by goguy123

Got a Prelim on monday

, please help, thanks

Q: Find the equation of the tangent to the parabola with equation
y=6+x-x(squared) at the point (1,6 ) .

If possible please explain to me how to do this question thanks .

Since you did not post this question in the Calculus subforum I assume a non-calculus approach is required ....?

Since the point (1, 6) lies on the tangent line, the equation of the tangent is $y - 6 = m(x - 1) \Rightarrow y = mx - m + 6$ where m is the gradient.

Also, the tangent (by definition) only intersects the parabola once. So the equations:

$y = 6 + x - x^2$ .... (1)

$y = mx - m + 6$ .... (2)

have only one solution. Solving (1) and (2) simultaneously:

$6 + x - x^2 = mx - m + 6 \Rightarrow x^2 + (m - 1)x - m = 0$ .

Since there's only one solution, the discriminant is equal to zero:

$(m - 1)^2 + 4m = 0$ .

Solve this equation for m and substitute the value into $y = mx - m + 6$ to get the final equation of the tangent.

Thread: Equation of tangent to parabola

LinkBack

Thread Tools

Display

Equation of tangent to parabola

Similar Math Help Forum Discussions

Find an Equation of the Tangent Line to the parabola at the given point and find....

Find the equation of the tangent to a parabola

parabola tangent

tangent of a parabola

Finding Equation of a Parabola from Tangent Slopes

Search Tags