Question

Two common tangents to the circle x^2+y^2=2a^2 and parabola y^2=8ax are

Answer 1

Answer:

Step-by-step explanation in the attachment

Answer 2

Two common tangents to the circle $x^{2} +y^{2} =2a^{2}$ and parabola $y^{2}=8ax$ are y = ±(x + 2a)

• Gvien parabola is

                   $y^{2}=8ax$     ---------------(1)

• Equation of tangent to (1) is

                   $y=mx+\frac{2a}{m}$   -----(2)

                   $y=\frac{m^{2} x+2a}{m} \\my=m^{2}x+2a\\ m^{2}x-my+2a=0$---------(3)

• Given circle is

                   $x^{2} +y^{2} =2a^{2} \\Center = (0, 0), Radius(r)=a\sqrt{2}$

• (3) is tangent to circle also

        ⇒ Perpendicular distance from (0, 0) to (3) = radius

             $\frac{|ax_{1}+by_{1}+c|}{\sqrt{a^{2}+b^{2}}}=r\\(x_{1}, y_{1})=(0,0)\\a=m^{2}, b=-m, c=2a$

             $\frac{|m^{2}(0)-m(0)+2a| }{\sqrt{m^{4}+m^{2}}}=a\sqrt{2} \\2=\sqrt{2(m^{4}+m^{2})}$

             Squaring on both sides

             $4=2(m^{4}+m^{2})\\m^{4}+m^{2}-2=0\\m^{4}+2m^{2}-m^{2} -2=0\\m^{2}(m^{2}+2)-(m^{2}+2)=0\\(m^{2}+2)(m^{2}-1)=0\\ m^{2}+2=0, m^{2}-1=0\\ m^{2}=-2(not possible), m^{2}=1$

               ∴ m = ±1

            Substituting m in (2)

              y = (±1)x+(2a)/(±1)

              y = ±(x+2a) are the two common tangents to the given circle and parabola