Determine the equation of the normal line
to y = x squared - 3x + 4
that is also parallel to 4x + 8y = 1.
Remember a normal line to a curve is perpendicular
to the tangent line at the point of tangent Z.
So if we take a look at this graph below,
the green line is a line given by 4x + 8y = 1.
So we're looking for the normal line to the curve
that is parallel to this green line.
And if this green line is parallel to the normal line,
it would also be perpendicular to the tangent line.
So for example,
if I sketch the tangent line at this red point,
notice how it's perpendicular to the tangent line.
And if I sketch the normal line,
which is in perpendicular to the tangent line
passing through the point of tangency,
it's also perpendicular to the tangent line.
To find the equation of the line normal to the curve
and also parallel to the given line,
we will have to find this point of tangency.
To do this, we'll have to find the derivative
of y = x squared - 3x + 4.
Considered equal to the slope of the tangent line,
which would be the negative reciprocal
of the slope of the line given by 4x + 8y = 1,
again because the two lines are perpendicular.
So let's first determine the slope of the given line
by solving 4x + 8y = 1 for Y.
First step, subtract 4x on both sides,
which gives us 8y = -4x + 1,
and then divide both sides by 8.
Simplifying, we have y = -1/2x + 1/8th.
Notice the slope of the green line
or the given line is negative 1/2.
Negative 1/2 is also going to be the slope
of the normal line to the curve
that's also parallel to the given line.
So m = 1/2 is the slope of the normal line.
And now as we're going to determine the point of tangency,
so again, to find that point of tangency,
we'll have to set the first derivative
equal to the slope of the tangent line.
The slope of the tangent line is equal
to the negative reciprocal of negative 1/2,
which is positive two over one or positive two.
So m = 2 is the slope of the tangent line.
And because the derivative function values
of y = x squared -3x + 4
give us these slopes of the tangent lines,
we will set the derivative
of y = x squared - 3x + 4 equal to 2,
and then solve for X.
This will give us the X coordinate of the point of tangency,
which we need to find the equation of the normal line.
So Y prime is equal to the derivative
of x squared - 3x + 4, which is 2X - 3.
And now we'll set this equal to the slope of 2
and solve for X.
Next, we add three to both sides which gives us 2x = 5.
Divide both sides by two.
Simplifying, we have x = 5/2.
So the point of tangency
which is also a point on the normal line we're looking for,
has an X coordinate of 5/2.
The Y coordinate is Y of 5/2,
which means we now need to substitute x = 5/2
back into the original equation or function
to determine the white coordinate.
So let's go ahead and do that.
Y of 5/2 is equal to the square of 5/2 - 3 x 5/2 + 4.
Y of 5/2 is equal to 11/4th.
Now we know the point of tangency
is the point 5/2 comma 11/4th.
Now we have all the information we need to find the equation
of the normal line to the given curve
that's also parallel to the given line.
We know the slope of the normal line is -1/2
and we know it passes through the point 5/2 comma 11/4th.
Let's first find the equation in point slope form
where x1 is 5/2 and y1 is 11/4
We have y - 11/4 = -1/2 X the quantity X minus 5/2.
This is the equation of the normal line
that we're looking for,
plus also about in slope-intercept form by solving for Y.
So the next step we'll distribute -1/2,
which gives us Y - 11/4 = -1/2x,
and then plus 5/4.
Next here, we'll add 11/4 to both sides.
Simplifying, we have y = -1/2x,
and then we have plus 16/4, which simplifies to 4.
Let's go ahead and verify this graphically.
We have the graph of the parabola.
The green line is the given line
and we just found the equation, the red line,
which is normal to the curve
and also parallel to the given line.
And again, if we sketch a tangent line at this red point,
notice how the normal line is perpendicular
to this tangent line,
and again, also parallel to the given green line.
I hope you found this helpful.