Vikas TU
Last Activity: 4 Years ago
16x^2−25y^2 =400
tangents from (2sqrt(2),1) is y=mx+c
c=1−2msqrt(2)
tangent to the hyperbola in slope form is
y=mx± sqrt(a^2m^2 −b^2)
c=1−2msqrt2
square both side
1+8m^2−4sqrt2m=25m^2−16
17m^2+4sqrt(2)m−17=0
from above equation we will get slope of tangents m1,m2
but before solving this we can see that m1m2 =−1
it means tangents are perpendicular to each other.
