Question

Ellipse!

Answer 1

?

Answer 2

From centre O of ellipse
x^2/16+ y^2/9=1 two perpendicular rays are drawn meeting ellipse at P and Q.N is foot of perpendicular from O to PQ, then

Answer 3

ON=?

Answer 4

what does N is foot of perpendicular from O to PQ mean ?

Answer 5

Draw a perpendicular from O to lime PQ.

Answer 6

is there anyway I can see the actual question
I'm assuming P and Q are points on the ellipse
I have no idea what N is...
Is N a point lying on a line segment that connect P and Q?

Answer 7

Yes!

Answer 8

Can we see diagrams please! :)

Answer 9

There is literally no diagram for this in my sheet.

Answer 10

@ganeshie8 sir pls help in this.

Answer 11

Draw your own diagram :) It will helps very much

Answer 12

I can't be able to becoz i use os on mobile.

Answer 13

Can you please tell me where are you finding difficulty in drawing this though?

Answer 14

so what is your actual question?

Answer 15

I think i wrote it just after darth's post.

Answer 16

Here's diagram to begin with

Answer 17

Correct!

Answer 18

Let P and Q be represented parametrically P(4 cos theta, 3sin theta) and with phi for Q
I recommend you find gradient of OP and OQ

Answer 19

I know you will say that
(bsin phi)(bsin theta)/(acos phi)(a cos theta)=-1

Answer 20

a=4,b=3

Answer 21

Can you just find out the length of ON just by using this ?

Answer 22

maybe we could use perpendicular distance formula

Answer 23

you'll need to write an eqn for the formula of the chord PQ first

Answer 24

Well, i tried that .
It came out to be y-bsintheta=bsintheta/acostheta(x-acostheta)

Answer 25

for chord PQ? I don't think so

Answer 26

Anyhow I'll show you the way with that.

The gradient of chord PQ is given by mPQ =
[bsin(theta) - bsin(phi)]/[acos(theta)-acos(phi)]

Answer 27

Oops! Typo .

Answer 28

you then use y - bsin(theta) = mPQ(x - asin(theta)
Rearranged this becomes mPQx - y + bsin(theta) - MPQ asin(theta) = 0
We can sub in 0,0 in x and y (as our perpendicular distance formula dictates) and the numerator becomes |bsin(theta) - MPQ asin(theta)| = |sin(theta)(b - aMPQ)|

Answer 29

it's quite algebra heavy, will take me a while to do. there is one other way is to write distances using similarity of right angled triangles. so ON^2 = OP^2 x OQ^2 / PQ^2
replace b^2 with a^2(1-e^2) when you can and with some rigourous working you should be able to arrive at a result.