Question

Find the coordinates of a point of the parabola $y=x^2+7x+2$ which is closest to the straight line $y=3x-3$.

Answer 1

Let the coordinates of the required point be $(h,k)$

Given parabola $x^2+7x+2$

$\therefore k=h^2+7h+2$

Using point - line distance formula for the line $3x-3$, we get,

$D=\frac{|3h-k-3|}{\sqrt{3^2+(-1{)}^2}}$

using the value of k, we get,

$D=\frac{|3h-h^2-7h-2-3|}{\sqrt{10}}$

$\therefore D=\frac{h^2+4h+5}{\sqrt{10}}$

upon differentiation, we get,

$\frac{dD}{dh}=\frac{2h+4}{\sqrt{10}}$

for maxima and minima, we have,

$\frac{dD}{dh}=0$

$\Rightarrow \frac{2h+4}{\sqrt{10}}=0$

upon simplification, we get,

$h=-2$

Now,

$\frac{d^{2}D}{d{h}^2}=\frac{2}{\sqrt{10}}>0$

$\therefore h=-2$ is the point of minima

substituting the value of h to obtain k, we get,

$k=(-2{)}^2+7(2)+2$

$k=-8$

Therefore, point closest to the parabola is $(-2,-8)$