Question

Question #9a3da

Answer 1

`(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:

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

Answer 2

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.