last question on this paper, C1

You need to find the equation of the normal then let

y = 0

to find where it intersects with the

x

axis.

(edited 13 years ago)

Reply 4

Original post by Femto

You need to find the equation of the normal then let y = 0 to find where it intersects with the x axis.

For the equation of the normal, I got y = (1/4)kx + 3k

Wtf?

Reply 5

I've got:

y=-2/3 x
as it is a normal it will be
y=-1/(-2/3) x
this equals to
y=-3/2 x
i.e.
y=-3x/2

what do I do next?

Reply 6

Original post by im so academic

For the equation of the normal, I got y = (1/4)kx + 3k

Wtf?

Yes because the preceding part of the question asks you to find

k

which I believe is 6 in this case.

Reply 7

Ok if

k = 6

then we can assume the following:

y = 6\sqrt x

Since the line

y = -\frac{2}{3}x

is parallel to the normal at the point which

x = 4

then the equation of the normal can be worked out since we have the point

P (4, 12)

and a gradient of

-\frac{2}{3}

.

I think you can carry on from there.

Reply 8

Farhan.Hanif93

18

Original post by Femto

Yes because the preceding part of the question asks you to find

k

which I believe is 6 in this case.

Original post by im so academic

I've got:

y=-2/3 x
as it is a normal it will be
y=-1/(-2/3) x
this equals to
y=-3/2 x
i.e.
y=-3x/2

what do I do next?

Original post by cooldudeman

i understand part one because someone has already helped me with it but part two...
i dont know how to get the coordinates for point Q and yes i know that y=0 on it but i dont know how to get the x coordinate.

You don't need to find the equation of the normal. You only need to find the gradient of the normal at x=4 in terms of k using differentiation. Then note that parallel lines have equal gradients. You know what the gradient of 2x+3y=0 is so equate the two expressions and solve for k.
k is 6.

for (ii) find the x-coordinate of Q and the y-coordinate of P and then use

A=\frac{1}{2}base\times height

.

(edited 13 years ago)

Reply 9

Original post by im so academic

I've got:

y=-2/3 x
as it is a normal it will be
y=-1/(-2/3) x
this equals to
y=-3/2 x
i.e.
y=-3x/2

what do I do next?

Which part are you asking about?

Reply 10

Original post by Farhan.Hanif93

You don't need to find the equation of the normal. You only need to find the gradient of the normal at x=4 in terms of k using differentiation. Then note that parallel lines have equal gradients. You know what the gradient of 2x+3y=0 is so equate the two expressions and solve for k.
k is 6.

Found the gradient at P = 1/4k
Therefore the normal will be 4k????
But the other line has a gradient of -3/2
How is it 6?

Reply 11

Original post by Femto

Which part are you asking about?

Trying to work out the value of k.

Reply 12

Original post by im so academic

Found the gradient at P = 1/4k
Therefore the normal will be 4k????
But the other line has a gradient of -3/2
How is it 6?

Well

y = k\sqrt x

right? If the normal is parallel to

2x + 3y = 0

then the tangent gradient is needed; this will be

\frac{3}{2}

We are given that

x = 4

at point P so if we differentiate

y = k \sqrt x

:

\frac{dy}{dx} = \frac{k}{2\sqrt x}

Set

\frac{dy}{dx} = \frac{3}{2}

and

x = 4

We get:

\frac{k}{4} = \frac{3}{2}

\rightarrow k = \frac{12}{2}

Which means

k = 6

(I've only explained the first part since OP said he knows how to do it.)

(edited 13 years ago)

Reply 13

OK this is what I've done now.

Equation of gradient at normal is y = 4kx
Equation of the other line is y = -2/3x

Equalise:

4kx = -2/3x

Cancel out x's:

4k = -2/3

Rearrange for K:

k = -1/6

What?

Reply 14

Farhan.Hanif93

18

Original post by im so academic

Found the gradient at P = 1/4k
Therefore the normal will be 4k????

The gradient of the normal at P is not 1/4k. I'm not sure what you mean by the normal being '4k'?...

y=k\sqrt{x} \implies \frac{dy}{dx}=\frac{k}{2\sqrt x}

When x=4,

\frac{dy}{dx}=\frac{k}{4}

.

This is the gradient of the TANGENT at P. You know that tangents and normals at a point are perpendicular to each other. From this, what can you tell me about the gradient of the normal at that point?

You know that this normal is parallel to 2x+3y=0 i.e. it has the same gradient. What is the gradient of this line? How does it relate to the gradient above?

Reply 15

Original post by Femto

Wait why is the tangent gradient needed? Doesn't normal = perpendicular?

Anyways, surely if you rearrange that equation for y you get -2/3?

Reply 16