Question

find the coordinate of a point on the parabola y=x^2 +7x+2 which is closest to the straight line y=3x-3?

Answer 1

first of all we should resolve the equation of parabola ,
e.g., y = x² + 7x + 2 = x² + 7x + (7/2)² + 2 - (7/2)²
y = x² + 2.(7/2)x + (7/2)² + 2 - 49/4
y = (x + 7/2)² - 41/4
now, we can let the point on the parabola .e.g., {(t - 7/2), t² - 41/4 }

Now, A/C to question, point perpendicular distance from the point { (t - 7/2) , (t² - 49/4)} to line y = 3x - 3 is shortest { Let P }

P = |3(t - 7/2) - (t² - 41/4) - 3 |/√{3² + 1²}
P = |3t - 21/2 - t² + 41/4 - 3 |/√10
now, differentiate P with respect to t { applying maxima and minima concept }
dP/dt = |3 - 0 - 2t + 0 - 0 |/√10
⇒dp/dt = 0
⇒ 3 - 2t = 0 ⇒t = 3/2

at t = 3/2 , { (t - 7/2) , (t² - 41/9) } ≡ { (3/2 - 7/2) , ( 9/4 - 41/4) } ≡(-2 , - 8)

Hence, closest point on the parabola y = x² + 7x + 2 which is closest to the straight line y = 3x - 3 is (-2,-8)

Answer 2

Hello sir,


Here is your answer,


Hope it helps you,
Thank you. :)