Question

How do you graph $4{x^2} + 49{y^2} + 294y + 245 = 0?$

Answer 1

Hint : To draw the graph of the given equation, compare it with the standard conic equation, you will get that the given equation is the equation of an ellipse. Then express the equation in standard form of an ellipse and then find coordinates of its center, vertices, co-vertices and foci and also find its eccentricity. And then with help of these parameters draw the required ellipse.

Complete step-by-step answer :
In order to draw the graph of the equation $4{x^2} + 49{y^2} + 294y + 245 = 0$ we will first find out what this equation represents by comparing it with the general form of conic equation,
Since the given equation has a squared form of both variables that is $x\;{\text{and}}\;y$ and also it has their coefficients positive, therefore the given equation is the equation of an ellipse.
Now converting the given equation into standard form of ellipse, we will get
$\Rightarrow 4{x^2} + 49{y^2} + 294y + 245 = 0 \\\ \Rightarrow 4{x^2} + {(7y)^2} + 2 \times 7y \times 21 + {21^2} - {21^2} + 245 = 0 \\\ \Rightarrow 4{x^2} + {(7y + 21)^2} - 441 + 245 = 0 \\\ \Rightarrow 4{x^2} + {7^2}{(y + 3)^2} - 196 = 0 \\\ \Rightarrow 4{x^2} + 49{(y + 3)^2} = 196 \;$
Now dividing both sides of the equation with $4 \times 49$ we will get

\Rightarrow \dfrac{{4{x^2} + 49{{(y + 3)}^2}}}{{4 \times 49}} = \dfrac{{196}}{{4 \times 49}} \\\ \Rightarrow \dfrac{{{x^2}}}{{49}} + \dfrac{{{{(y + 3)}^2}}}{4} = 1 \\\ \Rightarrow {\left( {\dfrac{{x - 0}}{7}} \right)^2} + {\left( {\dfrac{{y - ( - 3)}}{2}} \right)^2} = 1 $$ Comparing it with standard equation of ellipse $${\left( {\dfrac{{x - h}}{a}} \right)^2} + {\left( {\dfrac{{y - k}}{b}} \right)^2} = 1$$, we get $ h = 0,\;k = - 3,\;a = 7\;{\text{and}}\;b = 2 $ Now, finding parameters of the ellipse, Coordinates of center point: $ (h,\;k) \equiv (0,\; - 3) $ Coordinates of vertices: $ (h - a,\;k) \equiv (0 - 7,\; - 3) \equiv ( - 7,\; - 3)\;{\text{and}}\;(h + a,\;k) \equiv (0 + 7,\; - 3) \equiv (7,\; - 3) $ Coordinates of co-vertices: $ (h,\;k - b) \equiv (0,\; - 3 - 2) \equiv (0,\; - 5)\;{\text{and}}\;(h,\;k + b) \equiv (0,\; - 3 + 2) \equiv (0,\; - 1) $ Coordinates of foci: $$\left( {h - \sqrt {{a^2} - {b^2}} ,\;k} \right) \equiv \left( {0 - \sqrt {{7^2} - {2^2}} ,\; - 3} \right) \equiv \left( { - \sqrt {45} ,\; - 3} \right)\;{\text{and}}\;\left( {h + \sqrt {{a^2} - {b^2}} ,\;k} \right) \equiv \left( {0 + \sqrt {{7^2} - {2^2}} ,\; - 3} \right) \equiv \left( {\sqrt {45} ,\; - 3} \right)$$ Eccentricity of the ellipse will be given as $ \dfrac{{\sqrt {{a^2} - {b^2}} }}{a} = \dfrac{{\sqrt {45} }}{7} $ Now with help of above parameters, graphing the equation of the ellipse below  Therefore this the required graph of the given equation. **Note** : To graph this type of equations, your first step should be to check which conic section does the given equation belong to with help of either seeing the coefficients and powers of variables in the given equation or going through the determinant method. Also check whether it is horizontal or vertical conics.