Question

Find the equation of tangent and normal at(3, 2) of the circle x + y2-x-3y-430​

Answer 1

Answer:

We have, x2+y2-x-3y-430=0

Differentiating w.r.t x, we get

2x+2y(dy/dx)-1-3(dy/dx)=0

or, dy/dx = ((1-2x)/(2y-3))

Hence slope of tangent at point (3,2) is ((1-2.3)/(2.2-3)) i.e -5. Since tangent is perpendicular to normal and vice versa, slope of normal becomes 1/5.

Hence, equation of tangent at (3,2) is y-2=(-5)(x-3) i.e 5x+y-17=0 and equation of normal is y-2=(1/5)(x-3) i.e x-5y+7=0