Two common tangents to the circle x²+y²=2a² and parabola y²=8ax are

Correct Answer is : y=±(x+2a)y=± (x+2a)y=±(x+2a)

Correct Answer is : y=±(x+2a)y=± (x+2a)y=±(x+2a)

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Manipal Entrance Test 2015 Math Paper

Question : 62 of 70

Marks: +1, -0

x^{2}+y^{2}=2a^{2}

y^{2}=8ax

are

Solution:

The equation of any tangent to

y^{2}=8ax

y=mx+\frac{2a}{m}

...(i)

If it touches

x^{2}+y^{2}=2a^{2},

then

\left(\frac{2a}{m}\right)^{2}=2a^{2}(1+m^{2})

\Rightarrow 2=m^{2}(m^{2}+1)

\Rightarrow m^{4}+m^{2}-2=0

\Rightarrow (m^{2}+2)(m^{2}-1)=0

\Rightarrow m^{2}-1=0

\Rightarrow m=\pm 1

we get Putting the values of

m

in Eq. (i), we get

y=\pm (x+2a)

as the equations of common tangents.

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