The point on the curve $ y=x^{2} $ which is closest to $ \left(4,-\frac{1}{2}\right) $, is

Question

The point on the curve $ y=x^{2} $ which is closest to $ \left(4,-\frac{1}{2}\right) $, is, $ (1,1) $, $ (2,4) $, $ \left(\frac{2}{3}, \frac{4}{9}\right) $, $ \left(\frac{4}{3}, \frac{16}{9}\right) $

MARKS App · Accepted Answer

y = x 2 Let any point on this parabola is t , t 2 Equation of normal at this point is x + 2 t y = t + 2 t 3 It passes through 4, - 1 2 4 - t = t + 2 t 3 2 t 3 + 2 t - 4 = 0 t 3 + t - 2 = 0 t = 1 So point is 1 , 1

Search any question & find its solution

Looking for more such questions to practice?