Question

can anyone help me with this question..?
on a parabola there is a point whose abscissa and ordinate are equal; prove that, the normal chord at this point subtends a right angle at the focus..

Answer 1

I say you put "it's irrelevant to my life goals and ambitions" in the answer box. Scream and run out of the room.

Answer 2

abscissa and ordinate are just fancy names for x and y

Answer 3

i know that.. i am a bit confused regarding normal chord but i guess its the chord perpendicular to axis of parabola.. I have tried to prove what is asked in question but haven't succeed yet..

Answer 4

try maybe to express the parabola in parametric form....

Answer 5

is that point (1,1)

Answer 6

no idea.. but how do you get that point..?? we don't even have the equation of parabola.. i guess we have to do the whole thing based on variables..

Answer 7

\[x=at ^{2},y=2at\]
this are parametric equations of the parabola with focus at (a,0)
i think this would be a good start

Answer 8

switch x and y and you get parabola with focus at (0,a)

Answer 9

yeah, i know about parametric functions.. let me see.. can you also please try it for me..

Answer 10

At the bottom

http://www.askiitians.com/iit-jee-coordinate-geometry/normal-to-a-parabola.aspx

Answer 11

doesnt 1^2 = 1 ?

Answer 12

thanks @estudier
and @amistre64 i didn't get what you are trying to say..

Answer 13

the geometric definition of a parabola, i believe, is (x-h)^2 = 4a(y-k)

for any generic parabola; using x and y for ordinate and abiscuss

Answer 14

looks like estudiers got a nice link :)

Answer 15

yeah that's true.. but here we use eqn y^2=4ax and parametric points (at^2 , 2at)

Answer 16

I think i didn't get the question right ..