Question

Find coordinates of points on parabola $y=x^2+x+1$ which is closest to $y=(x-1)$ ?

Answer 1

First of all we should resolve the equation of parabola ,

e.g., $y=x²+7x+2=x²+7x+\left(7/2\right)²+2-\left(7/2\right)²$

$y=x²+2.\left(7/2\right)x+\left(7/2\right)²+2-49/4$

$y=(x+7/2)²-41/4$

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 $LetP$

$P=|3(t-7/2)-(t²-41/4)-3|/\surd 3²+1²$

$P=|3t-21/2-t²+41/4-3|/\surd 10$

now, differentiate P with respect to t { applying maximam and minimam concept }

$dP/dt=|3-0-2t+0-0|/\surd 10$

$dp/dt=0$

$3-2t=0\Rightarrow t=3/2$

at $t=3/2,(t-7/2),(t²-41/9)\equiv (3/2-7/2),(9/4-41/4)\equiv (-2,-8)$

Hence, closest point on the parabola $y=x²+7x+2$ which is closest to