# Find the equation of the straight line which is tangent at one point and normal at another point on the curve x=3t^2 , y=2t^3

Y RAJYALAKSHMI
45 Points
9 years ago
The slope of the tangent at P(t) is dy/dx = dy/dt * dt/dx = t
Let the tangent cuts the curve and normal at Q(t1), then the coordinates of the point is (3t12, 2t13)
The the slope of the line passing throught the points P & Q  is (2t1- 2t3)/ (3t12 – 3t2) = t
Solving this equation we get, t1 = – t/2
Then the coordinates of the Q = (3t2/4, – t3/4)
Slope of the normal at Q is – dx/dy = 2/t
The equation of the straight line having slope t and passing throught Q is y – (– t3/4) = t (x – 3t2/4)
=> tx – y = t3 ------- (1)
The equation of the straight line having slope 2/t and passing throught Q  is y – (– t3/4) = 2/t (x – 3t2/4)
=> 2x/t – y = 3t/2 + t3/4  ----------- (2)

Since these two equations represent the same straight line, by comparing the corresponding coefficients, we have t = sqrt (2)

By substituting this value in any of the above equaitons, we get sqrt(2) * x – y = 2 sqrt(2)

The equation of the required straight line is sqrt(2) * x – y = 2 sqrt(2)