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Q.Show that the equation of common tangents to the circle x2+y2=2a2x^2+y^2=2a^2 and the parabola y2=8axy^2=8ax are y=±(x+2a)y=\pm(x+2a).

Telangana TsbieTelangana Board of Intermediate Education 2019Subjective· 7mImportance★★★★★
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Write the tangent to the parabola in slope form y=mx+2amy=mx+\frac{2a}{m}, then force it to also be tangent to the circle by equating the perpendicular distance from the centre to the radius.

Parabola: y2=8ax=4(2a)xy^2=8ax=4(2a)x, so comparing with y2=4Axy^2=4Ax we get A=2aA=2a.

A tangent to y2=4Axy^2=4Ax with slope mm is y=mx+Am=mx+2amy=mx+\dfrac{A}{m}=mx+\dfrac{2a}{m}.

Circle: x2+y2=2a2x^2+y^2=2a^2, centre (0,0)(0,0), radius r=a2r=a\sqrt2.

For this line to also be tangent to the circle, the perpendicular distance from (0,0)(0,0) to mx−y+2am=0mx-y+\dfrac{2a}{m}=0 must equal rr:

∣2am∣m2+1=a2\dfrac{\left|\dfrac{2a}{m}\right|}{\sqrt{m^2+1}}=a\sqrt2

Square both sides:

4a2m2=2a2(m2+1)\dfrac{4a^2}{m^2}=2a^2(m^2+1)

Divide by 2a22a^2:

2m2=m2+1\dfrac{2}{m^2}=m^2+1

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