Question

# If the tangent to the ellipse $$x^2+4y^2=16$$ at the point $$P(\phi)$$ is a normal to the circle $$x^2+y^2-8x-4y=0$$ then $$\phi$$ is equal to

A
π2
B
π4
C
tan112
D
π4

Solution

## The correct option is B $$tan^{-1}\frac{1}{2}$$$$y=mx+\sqrt {16m^2-4}$$$$c=\pm \sqrt {a^2m^2+b^2}$$$$\sqrt {16m^2+4}$$Now it is normal to circle so it passes through centre$$\therefore (4, 2)$$$$2=4m \pm \sqrt {16m^2-4}$$$$1=2m\pm \sqrt {4m^2-1}$$$$(1-2m)^2=(4m^2-1)$$$$4m^2-4m+1=4m^2-1$$$$4m=2, m=\frac {1}{2}$$$$\phi =tan^{-1}\frac {1}{2}$$Maths

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