The equation of a common tangent tangent to the curves, `y^(2)=16x and xy= -4,` is

Options

A. `x-y+4=0`
B. `x+y+4=0`
C. `x-2y+16=0`
D. `2x-y+2=0`

Explanation

Key Idea An equation of tangent having slope
'm' to parabola `y^(2)=4ax " is " y=mx+(a)/(m)`.
Given equation of curves are
`y^(2)` =16x(parabola)` " …(i)" `
and xy = -4 (rectangular hyperbola) ` " …(ii)" `
Clearly, equation of tangent having slope 'm' to parabola
(i) is `y=mx+(4)/(m) " ...(iii)" `
Now, eliminating y from Eqs. (ii) adn (iii), we get
` x (mx+(4)/(m))= -4 rArr mx^(2)+(4)/(m)x+4=0.`
which will give the points of intersection of tangent and rectangular hyperbola.
Since, line `y=mx+(4)/(m) ` is also a tangent to the rectangualr hyperbola.
` therefore ` Discriminant of quadratic equation `mx^(2) +(4)/(m)x+4=0,`
should be zero.
[ `therefore ` there will be only one point of intersection ]
`rArr D=((4)/(m))^(2)- 4(m)(4)=0`
`implies m^(3)=1 rArr m=1`
So, equation of required tangent is `y=x+4.`