1 Year on Brilliant!

Geometry Level 5

Tangent is drawn at any point ( p , q ) (p,q) on the parabola y 2 = 4 a x y^2=4ax . Tangents are drawn from any point on this tangent to the circle x 2 + y 2 = a 2 x^2+y^2=a^2 , such that the chords of contact pass through a fixed point ( r , s ) (r,s) . Then p , q , r , s p,q,r,s hold which of the given relations?

r 2 q = 4 p 2 s r^2q=4p^2s r 2 q = 4 p 2 s r^2q=-4p^2s r q 2 = 4 p s 2 rq^2=4ps^2 r q 2 = 4 p s 2 rq^2=-4ps^2

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

We know tangent to a conic is given by T=0 ,

i.e. Tangent to parabola at ( p , q ) (p,q) is

2 a x y q = 2 a p 2ax -yq=-2ap .

Substitute x and y in this equation by ( x 1 , y 1 ) (x_1 , y_1) ,

Where ( x 1 , y 1 ) (x_1 , y_1) are points where tangent

2 a x 1 y 1 q = 2 a p 2ax_1 -y_1 q=-2ap .

Call this as equation 1 1 .

Now equation of chord of contact of circle is T=0,passing through ( r , s ) (r,s) and ( x 1 , y 1 ) (x_1 , y_1) .

Therefore

r x 1 + s y 1 = a 2 rx_1 + sy_1=a^{2} . ……… 2 2 .

Since 1 1 and 2 2 are identical (in x 1 x_1 and y 1 y_1 ,

r / 2 a = s / q = a / 2 p r/2a=-s/q=-a/2p .

Now let each of these ratios be ξ \xi .

Now we get

a = 2 p ξ a=-2p \xi

Since r = 2 a ξ = 4 p ( ξ ) 2 r= 2a \xi = -4p( \xi)^{2}

And s = q ξ s=-q \xi .

Eliminating ξ \xi we get

r q 2 = 4 p s 2 \boxed{rq^{2}=-4ps^{2}} .

Must be a level 3 or 4 problem

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...