How do you find the equations of both lines through point (2,-3) that are tangent to the parabola #y=x^2+x#?

2 Answers
Feb 6, 2017

Answer:

The equations of the tangents that pass through #(2,-3)# are:

# y=-x-1 # and
#y = 11x-25#

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. So for our curve (the parabola) we have

# y=x^2+x #

Differentiating wrt #x# we get:

# dy/dx=2x+1 #

Let #P(alpha,beta)# be any generic point on the curve. Then the gradient of the tangent at P is given by:

# m = 2alpha + 1 \ \ \ # (using the derivative)

And as P lies on the curve we also have:

# beta = alpha^2+alpha \ \ \ # (using the curve equation)

And so the tangent at #P# passes through #(alpha,alpha^2+alpha)# and has gradient #2alpha + 1#, so using the point/slope form #y−y_1=m(x−x_1)# the equation of the tangent at #P# is;

#y - (alpha^2+alpha) = (2alpha+1)(x-alpha)#

if this tangent also passes through #(2,-3)# then;

# \ \-3 - (alpha^2+alpha) = (2alpha+1)(2-alpha)#
# :. -3 - alpha^2-alpha = 3alpha-2alpha^2+2#
# :. alpha^2 -4alpha-5=0#
# :. (alpha-5)(alpha+1)=0#
# :. alpha =-1,5#

If #alpha =-1 => beta = 0 #, and the tangent equation becomes:

#y - 0 = (-1)(x+1)#
# :. \ \ y=-x-1 #

If #alpha =5 => beta = 30#, and the tangent equation becomes:

# \ \ \ \ \ y - 30 = (11)(x-5)#
# :. y - 30 = 11x-55#
# :. \ \ \ \ \ \ \ \ \y = 11x-25#

Hence the equations of the tangents that pass through #(2,-3)# are
# y=-x-1 # and #y = 11x-25#

We can confirm this graphically:
enter image source here

Feb 6, 2017

Answer:

The two tgts. are #11x-y-25=0, and, x+y+1=0#.

Explanation:

Let #m# be the slope of tgt. line thro. the pt. #(2,-3).#

Using the Slope-point Form , we get, its eqn., #y+3=m(x-2).#

Solving this eqn. with #y=x^2+x#, we get the Pts. of Intersection of

the Parabola with the tgt. line.

#x^2+x=y=m(x-2)-3=mx-2m-3, i.e.,#

# x^2+x(1-m)+(2m+3)=0..............(star).#

#(star)# being a quadratic eqn. in #x#, it can have at most #2# roots that may give the #x-#co-ords. of the Pts. of Int.

But, a tgt. essentially intersects the Parabola in only one pt., #(star)# must have #2# equal roots, meaning that, #Delta=0," for "(star).#

#rArr (1-m)^2=4(1)(2m+3)#

#rArr m^2-10m-11=0 rArr m=11, or, m=-1#.

For #m=11," the tgt. is "y+3=11(x-2)=11x-22, i.e., 11x-y-25=0,# and, If

#m=-1," the tgt. is, "x+y+1=0;# as Respected Steve has obtained!

N.B.:
#(1): m=-1,&, (star) rArr x^2+2x+1=0 rArr x=-1, so, y=x^2+x=0.#
#:." the pt. of contact is "(-1,1).#

#(2): m=11" gives "(5,30)" as the pt. of contact."#

Enjoy Maths.!