Video Transcript
Find the equation of the normal to
the curve π₯ equals four π squared minus six π and π¦ equals five π squared at π
equals one.
Letβs firstly remind ourselves what
the normal is. Letβs say we have a curve π¦ and we
want to find the normal to the curve at this point. We have a tangent to the curve
which has the same gradient dπ¦ by dπ₯ as the curve π¦. The normal is the line that is
perpendicular to the tangent at a point. And therefore, the normal has
gradient negative one over dπ¦ by dπ₯.
But what is the point that weβre
interested in for this question? Well, we have π₯ and π¦ defined in
terms of a third parameter, π. And we do know that we want the
equation of the normal at the point where π is equal to one. So, by substitution of π is equal
to one, we have that π₯ equals four multiplied by one squared minus six times one,
which gives us four minus six. And that gives us negative two. And by substituting π equals one
into π¦, we have that π¦ equals five multiplied by one squared, which just gives us
five.
So, we want the equation of the
normal at the point negative two, five. Now, we said that the gradient of
the normal is negative one over dπ¦ by dπ₯. If weβre able to find the gradient
of the normal, we can use this and the point weβve just found in order to find the
equation of the normal. So, we need to start by finding dπ¦
by dπ₯. We may often do this by
differentiating π¦ with respect to π₯. But weβve not been given π¦ in
terms of π₯, so we differentiate parametrically. This means that if π₯ and π¦ are
functions of a third parameter π, then dπ¦ by dπ₯ is equal to dπ¦ by dπ over dπ₯
by dπ. And thatβs okay as long as dπ₯ by
dπ is not equal to zero.
So, letβs start by finding dπ¦ by
dπ. This is the derivative of π¦ with
respect to π. π¦ equals five π squared. And using the power rule of
differentiation, which tells us to multiply by the power and then subtract one from
the power, we have that dπ¦ by dπ is equal to 10π. dπ₯ by dπ is the derivative of
π₯ with respect to π. As π₯ equals four π squared minus
six π, again using the power rule, we find that dπ₯ by dπ is equal to eight π
minus six. So, dπ¦ by dπ₯ is equal to 10π
over eight π minus six.
Remember that weβre only interested
in the gradient at a specific point. And that point is where π is equal
to one. So, by substitution of π is equal
to one, dπ¦ by dπ₯ equals 10 over two. And we can actually simplify this
because 10 over two is just five. Remember that we said that dπ₯ by
dπ, the denominator, must not be equal to zero. And because we found dπ₯ by dπ to
be equal to eight π minus six and at the point where π is equal to one, this is
just two, weβre okay to have used that formula.
Now, remember we said the normal
gradient is equal to negative one over dπ¦ by dπ₯. And because we found dπ¦ by dπ₯ to
be equal to five, then using this formula, we have that the gradient of the normal
is equal to negative one over five at the point where π is equal to one. So now, we have the gradient of the
normal and our point of interest negative two, five. We can find the equation of the
normal.
We know the point-slope form of the
general equation for a line. This is where we know the gradient
and a point on the line. And that formula is given by π¦
minus π¦ one equals π π₯ minus π₯ one, where π is the gradient and π₯ one, π¦ one
is a point on the line. So, with the substitution of π
equals negative one over five and π₯ one equals negative two and π¦ one equals five,
we have that the equation of the normal is π¦ minus five equals negative one over
five multiplied by π₯ minus negative two.
Of course, this is not a very nice
way to present our answer, so weβre going to do a little bit of simplifying
here. Firstly, in the brackets, we have
π₯ minus negative two. And we know when we subtract a
negative, this is the same as adding. So, this is the same as π₯ add
two. Letβs also multiply through by five
to get rid of this fraction here. This gives us five π¦ minus 25
equals negative π₯ minus two. This is the same as having a
negative one at the front of the brackets π₯ add two. So, letβs multiply this bracket
through by negative one.
From here, weβre just going to
rearrange to get our final answer. Weβll try to have all of our terms
on one side of the equals. So, letβs start by adding 25 to
both sides, which gives us five π¦ equals negative π₯ add 23. And then, weβll subtract five π¦
from both sides, which gives us our final answer of negative five π¦ minus π₯ add 23
equals zero.