The equation to the line touching both the parabolas `y^2 =4x` and `x^2=-32y` is

Options

A. `x+2y+4=0`
B. `2x+y-4=0`
C. `2x+y-4=0`
D. `x-2y+4=0`

Answer

C

Explanation

Any tangent to the parabola
`y^(2)=4x " is " y=mx+(1)/(m) ` solving this with `x^(2)=- 32 y`
`x^(2)=-32(mx+(1)/(m))`
`mx^(2)+32 m^(2) x +32 = 0`
`D=0 implies (32 m^(2))^(2) -4xxmxx32=0`
`therefore 32 m^(4)-4m=0`
` 8m^(3)- m=0`
` m=0,1//2,`
`m ne 0`
` therefore y=(x)/(2)+2`