Question #9a3da

2 Answers
Michael
Sep 23, 2015

#(a). 66.84^@#

#(b). (13/4,1/2)#

Explanation:

#x^2+y^2-y-3=0#

As you can see this describes a circle:

graph{x^2+y^2-y-3=0 [-10, 10, -5, 5]}

The 2 tangents where the line #x=1# cuts the circle fit the general equation of a straight line:

#y=mx+c#

We can find #m# and #c# to get the equations of the tangents and then find the angle of intersection.

To get the gradients, we can differentiate both sides implicitly since both #x# and #y# appear together:

#D(x^2+y^2-y-3)=D(0)#

#2x+2yy'-y'=0#

#y'(2y-1)=-2x#

#y'=(-2x)/(2y-1)#

If #x=1# then that line will cut the circle giving 2 values for #y#:

#1+y^2-y-3=0#

#y^2-y-2=0#

Factorising:

#(y+1)(y-2)=0#

If #(y+1)=0#

#y=-1#

If #(y-2)=0#

#y=2#

So these are the 2 values of #y# where the #x=1# line cuts the circle.

Now to get the equation of the 1st tangent where #x=1# and #y=-1#:

#m=y'=(-2x)/(2y-1)=(-2xx1)/(2xx(-1)-1)=(-2)/(-3)=2/3#

#m=2/3#

To get #c#:

#y=mx+c#:
#-1=2/3xx1+c#

#c=-1-2/3=-3/3-2/3=-5/3#

The equation for the tangent #rArr#

#y=2/3x-5/3" "color(red)((1))#

Now to get the 2nd tangent:

#x=1, y=2#

#m=y'=(-2x)/(2y-1)=(-2xx1)/((2xx2)-1)#

#m=-2/3#

To get #c#:

#y=mx+c#

#2=-2/3xx1+c#

#c=2+2/3=6/3+2/3=8/3#

The equation becomes:

#y=-2/3x+8/3" "color(red)((2))#

At the intersection of the 2 tangents we can put #color(red)((1))# equal to #color(red)((2))rArr#

#2/3x-5/3=-2/3x+8/3#

#(2x)/3+(2x)/3=8/3+5/3#

#(4x)/3=13/3#

#x=13/4#

Now to get the #y# co-ordinate:

#y=(2x)/3-5/3#

#y=2/3xx13/4-5/3#

#y=26/12-5/3=26/12-20/12=6/12=1/2#

#y=1/2#

So the (x,y) co-ordinates of the intersection are #(13/4,1/2)#

To get the angle of intersection between the 2 tangents there is an expression you can use but I think its best to look at the geometry to see what's going on:

MFDocs

Here you can see that the angle of intersection is #66.84^@#

Jim H
Oct 2, 2015

Michael has given a fine answer using a bit of calculus. Here is a partial answer without calculus.

Explanation:

The circle #x^2+y^2-y-3=0# can be put into standard form:

#x^2+(y-1/2)^2 = 13/4#

The center of the circle is #(0,1/2)#.

The points on the circle with #x=1# are #(1,2)# and #(1,-1)#

To find the slopes of the tangent lines at these points, use the fact that a tangent to a circle at a point is perpendicular to the radius at that point.

At #(1,-1)#, the slope of the radius is #(-1-(1/2))/(1-0) = -3/2#

So the slope of the tangent at #(1,-1)# is #2/3#
See Michael's answer to get the equation of the tangent line at #(1,-1)# is
#y=2/3x-5/3" "color(red)((1))#

At #(1,2)#, the slope of the radius is #(2-(1/2))/(1-0) = 3/2#

So the slope of the tangent at #(1,2)# is #-2/3#
See Michael's answer to get the equation of the tangent line at #(1,2)# is
#y=-2/3x+8/3" "color(red)((2))#

For the remainder of the solution, see Michael's answer.

Related questions
See all questions in Definition of the Tangent Line
Impact of this question
2247 views around the world
You can reuse this answer
Creative Commons License