Question

The minimum distance of 4x² + y + 4x – 4y + 5 = 0 from the line - 4x + 3y = 3 is​

Answer 1

Given:

An equation $4x^{2} + y^{2} + (4x) -(4y) + 5 = 0$ and a line $4x + 3y = 3$.

To Find:

The minimum distance of $4x^{2} + y^{2} + (4x) -(4y) + 5 = 0$ from the line $4x + 3y = 3$  is​ ?

Solution:

given equation is -

$4x^{2} + y^{2} + (4x)-( 4y) + 5 = 0\\(4x^{2} +1+4x)+(y^{2}+4-4y )=0\\(2x+1)^{2} +(y-2)^{2} =0\\4(x+\frac{1}{2} )^{2}+(y-2)^{2}=0$

and the line is-

$4x+3y=3\\4x+3y-3=0$

therefore, distance between given equation and line is -

$=4(\frac{-1}{2})+3(2)-3\\= -2+6-3\\=1$

Hence, minimum distance of $4x^{2} + y^{2} + (4x) -(4y) + 5 = 0$ from the line $4x+3y=3$ is 1.