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Miscellaneous Exercise 7(II) · Q125

Q.A line touches the circle x2+y2=2x^2 + y^2 = 2 and the parabola y2=8xy^2 = 8x. Show that its equation is y=±(x+2)y = \pm(x+2).

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y2=8x⇒a=2y^2=8x \Rightarrow a=2. A tangent to the parabola with slope mm is y=mx+2my=mx+\dfrac{2}{m}, i.e. c=2mc=\dfrac2m.

For this line to also touch the circle x2+y2=2x^2+y^2=2 (centre OO, radius 2\sqrt2), the perpendicular distance from OO to mx−y+c=0mx-y+c=0 must equal 2\sqrt2:

∣c∣m2+1=2⇒c2=2(m2+1).\dfrac{|c|}{\sqrt{m^2+1}}=\sqrt2 \Rightarrow c^2=2(m^2+1).

Substitute c=2mc=\dfrac2m: 4m2=2m2+2⇒4=2m4+2m2⇒m4+m2−2=0\dfrac{4}{m^2}=2m^2+2 \Rightarrow 4=2m^4+2m^2 \Rightarrow m^4+m^2-2=0. …

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