Question #9a3da

2 Answers
Sep 23, 2015

Answer:

#(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^@#

Oct 2, 2015

Answer:

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.