Question-225904

Question Number 225904 by ajfour last updated on 15/Nov/25

Commented by ajfour last updated on 15/Nov/25

If parabola is y=x^2 . Find equations of both these circles of equal radii.

$${If}\:{parabola}\:{is}\:{y}={x}^{\mathrm{2}} .\:{Find}\:{equations} \\ $$$${of}\:{both}\:{these}\:{circles}\:{of}\:{equal}\:{radii}. \\ $$

Answered by mr W last updated on 16/Nov/25

Commented by fantastic2 last updated on 16/Nov/25

from the picture we can?see Rsin θ=k...i 2Rsin θ=(k/2)+(k^2 /(sin θ)) ...ii ii/i 2=(1/2)+(k/(sin θ)) tan θ=2k sin θ=k/(√(1+4k^2 )) (k/k)(√(1+4k^2 ))=(3/2) 1+4k^2 =(9/4) 4k^2 =(5/4) k=((√5)/4) R=(k/(sin θ))=(k/k)(√(1+4k^2 ))=(3/2)✓

$${from}\:{the}\:{picture}\:{we}\:{can}?{see} \\ $$$${R}\mathrm{sin}\:\theta={k}…{i} \\ $$$$\mathrm{2}{R}\mathrm{sin}\:\theta=\frac{{k}}{\mathrm{2}}+\frac{{k}^{\mathrm{2}} }{\mathrm{sin}\:\theta}\:…{ii} \\ $$$${ii}/{i} \\ $$$$\mathrm{2}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{{k}}{\mathrm{sin}\:\theta} \\ $$$$\mathrm{tan}\:\theta=\mathrm{2}{k} \\ $$$$\mathrm{sin}\:\theta={k}/\sqrt{\mathrm{1}+\mathrm{4}{k}^{\mathrm{2}} } \\ $$$$\frac{{k}}{{k}}\sqrt{\mathrm{1}+\mathrm{4}{k}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{1}+\mathrm{4}{k}^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mathrm{4}{k}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${k}=\frac{\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${R}=\frac{{k}}{\mathrm{sin}\:\theta}=\frac{{k}}{{k}}\sqrt{\mathrm{1}+\mathrm{4}{k}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{2}}\checkmark \\ $$

Commented by mr W last updated on 16/Nov/25

y=(x^2 /c) (c=1 here) say both circles with radius R touch the parabola at (p, (p^2 /c)). tan θ=(dy/dx)=((2p)/c)=m, say ⇒p=((cm)/2) p=R sin θ ((mc)/2)=((Rm)/( (√(1+m^2 )))) ⇒m^2 =((4R^2 )/( c^2 ))−1 (p^2 /c)−R cos θ=R ((m^2 c)/4)−(R/( (√(1+m^2 ))))=R ((m^2 c)/4)−(c/( 2))=R (c/4)(((4R^2 )/c^2 )−1)−(c/( 2))=R R^2 −cR−((3c^2 )/4)=0 (R−((3c)/2))(R+(c/2))=0 ⇒R=((3c)/2)=(3/2)=1.5 for c=1

$${y}=\frac{{x}^{\mathrm{2}} }{{c}}\:\left({c}=\mathrm{1}\:{here}\right) \\ $$$${say}\:{both}\:{circles}\:{with}\:{radius}\:{R}\:{touch} \\ $$$${the}\:{parabola}\:{at}\:\left({p},\:\frac{{p}^{\mathrm{2}} }{{c}}\right). \\ $$$$\mathrm{tan}\:\theta=\frac{{dy}}{{dx}}=\frac{\mathrm{2}{p}}{{c}}={m},\:{say} \\ $$$$\Rightarrow{p}=\frac{{cm}}{\mathrm{2}} \\ $$$${p}={R}\:\mathrm{sin}\:\theta \\ $$$$\frac{{mc}}{\mathrm{2}}=\frac{{Rm}}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }} \\ $$$$\Rightarrow{m}^{\mathrm{2}} =\frac{\mathrm{4}{R}^{\mathrm{2}} }{\:{c}^{\mathrm{2}} }−\mathrm{1} \\ $$$$\frac{{p}^{\mathrm{2}} }{{c}}−{R}\:\mathrm{cos}\:\theta={R} \\ $$$$\frac{{m}^{\mathrm{2}} {c}}{\mathrm{4}}−\frac{{R}}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}={R} \\ $$$$\frac{{m}^{\mathrm{2}} {c}}{\mathrm{4}}−\frac{{c}}{\:\mathrm{2}}={R} \\ $$$$\frac{{c}}{\mathrm{4}}\left(\frac{\mathrm{4}{R}^{\mathrm{2}} }{{c}^{\mathrm{2}} }−\mathrm{1}\right)−\frac{{c}}{\:\mathrm{2}}={R} \\ $$$${R}^{\mathrm{2}} −{cR}−\frac{\mathrm{3}{c}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$$\left({R}−\frac{\mathrm{3}{c}}{\mathrm{2}}\right)\left({R}+\frac{{c}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow{R}=\frac{\mathrm{3}{c}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{1}.\mathrm{5}\:{for}\:{c}=\mathrm{1} \\ $$

Commented by mr W last updated on 16/Nov/25