express y in terms of x in the given equation 4y^2+24y-14-16x+4^2=0

Question

hence discuss the nature of the equation whether local minimum or local maximum

Answer 1

I guess this is supposed to be 4y^2+24y-14-16x+4x^2=0. This is the same as 2y^2+12y-7-8x+2x^2=0

Complete the square for y: 2(y^2+6y)-7-8x+2x^2=0; 2(y^2+6y+9)-2*9-7-8x+2x^2=0;

2(y+3)^2-25-8x+2x^2=0, so 2(y+3)^2=25+8x-2x^2.

(y+3)^2=12.5+4x-x^2; y+3=±√(12.5+4x-x^2) so y=-3±√(12.5+4x-x^2).

(y=-3±√(16.5-(x-2)^2) is another way of writing this.)

The coefficients of x^2 and y^2 are both 4 which means that the original equation graphs as a circle.

The centre of the circle is on the line y=-3. The max and min are on either side of the centre representing the highest and lowest points of the circle given by the value of the square root for a particular value of x (=2) defining the x coord for the centre of the circle, and so defining the radius (√16.5).