Circle Tangent to a Parabola?

Geometry Level 5

The above picture shows the graphs of the parabola y = x 2 y = x^{2} and the line y = 1 2 x y = -\frac{1}{2}x in Cartesian coordinates. A circle with center ( p , 0 ) (p, 0) , where p > 0 p > 0 , is tangent to both the parabola and the line. What is the value of p p rounded to three decimal places?


Graph generated in Desmos .


The answer is 0.677.

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1 solution

Ronak Agarwal
Oct 25, 2014

We have the centre of the circle as A = ( x , 0 ) A=(x,0) let the point of contact of circle and parabola be B = ( t , t 2 ) B=(t,{t}^{2}) .

Ist equation we would be writing is that the tangent and AB is perpendicular to each other, hence we write :

( S l o p e o f t a n g e n t ) ( S l o p e A B ) = 1 (Slope\quad of \quad tangent)(Slope\quad AB)=-1

( 2 t ) ( t 2 t x ) = 1 (2t)(\frac{{t}^{2}}{t-x})=-1

rewriting it we have :

2 t 3 + t = x 2{t}^{3}+t=x Equation 1

Second equation we would be writing is that :

length(AB)=Perpendicular distance from the line x + 2 y = 0 x+2y=0

t 4 + ( t x ) 2 = x 2 5 \Rightarrow {t}^{4}+{(t-x)}^{2}=\frac{{x}^{2}}{5} Equation 2

Solving these two equations we get :

x = ( 15 + 65 ) ( 2 65 2 ) 128 \large x=\frac{(15+\sqrt{65})(\sqrt{2\sqrt{65}-2})}{128}

Hello Ronak Agarwal, that's a very good solution. However, I also tried solving it using a different approach and I got the final answer as 830 + 130 65 64 \frac{\sqrt{830+130\sqrt{65}}}{64} which has the same value as your answer but in a different form. Can you please guide me on how to convert my answer into this form? ( a + b ) ( c + d e ) f \frac{(a + \sqrt{b})(\sqrt{-c+d\sqrt{e}})}{f} Thank you very much.

Daren Wong - 5 years, 6 months ago

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actually the question has been wrongly asked i did get same and also converted it but the answer i received was different check this (8+sqrt(65))(sqrt(-28130+3490sqrt(65))/64

aryan goyat - 5 years, 3 months ago

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