Parabolic Circle

Geometry Level 4

Consider the parabola y = a x 2 + b x + c y=ax^2 + bx+c , if it intersects x x -axis at α > 0 \alpha > 0 and β > 0 \beta > 0 , find the length of the tangent from the origin to the circle passing through α \alpha and β \beta when a = 2 , b < − 120 , c = 162 a=2, b<-120, c=162 . Consider α \alpha and β \beta to be diametric end points.


The answer is 9.

Prince Loomba by Prince Loomba , 21, India

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2 solutions

Ujjwal Rane
Dec 30, 2017

X-intercepts A and B of the parabola are solutions α \alpha and β \beta of the quadratic equation, a x 2 + b x + c = 0 ax^2 + bx + c = 0 where, α = − b − b 2 − 4 a c 2 a \alpha = \frac{-b - \sqrt{b^2-4ac}}{2a} and β = − b + b 2 − 4 a c 2 a \beta = \frac{-b + \sqrt{b^2-4ac}}{2a}

They are also diametrically opposite on the circle with center O C = α + β 2 OC = \frac{\alpha + \beta}{2} and radius C T = β − α 2 CT = \frac{\beta - \alpha}{2} Hence the desired length of tangent O T = O C 2 − C T 2 OT = \sqrt{OC^2 - CT^2}

Simplifying, O T = c a = 162 2 = 81 = 9 OT = \sqrt{\frac{c}{a}} = \sqrt{\frac{162}{2}} = \sqrt{81} = 9

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Prince Loomba
Apr 16, 2016

Length of tangent= S o r i g i n \sqrt {S_{origin}} . We know that circle's equation is ( x − α ) × ( x − β ) + y 2 = r 2 (x-\alpha)\times (x-\beta)+y^{2}=r^{2} . Length of tangent= α × β = c / a \sqrt {\alpha \times \beta}=\sqrt {c/a} . (By quadratic equation product of roots)=9

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Can alpha and beta both be positive when a, b, c are all positive? I think b > 120 is contradictory.

Ujjwal Rane - 3 years, 5 months ago

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