The equation of a tangent to the hyperbola `16x^2-25y^2-96x+100y-356=0`, which makes an angle `(pi)/(4)` with the transverse axis, is

Options

A. `y=x+2`
B. `y=x-5`
C. `y=x+3`
D. `x=y+2`

Explanation

The equation of the hyperbola is
`16(x^(2)-6x)-25(y^(2)-4y)=356implies((x-3)^(2))/(25)-((y-2)^(2))/(16)=1`
The equations of tangents of slope `m=tanpi//4=1` to this hyperbola are
`y-2-1xx(x-3)+-sqrt(25xx1^(2)-16)` [Using `y-k=m(x-h)+-sqrt(a^(2)m^(2)-b^(2))`]
`impliesy-2=(x-3)+-3impliesy=x+2` and `y=x-4`