Question

FIND THE FOCI FOR EACH EQUATION OF AN ELLIPSE.THEN GRAPH THE ELLIPSE
$49x ^{2} + y ^{2} = 49$​

Answer 1

Answer:

Tip: A regular oval shape, traced by a point moving in a plane so that the sum of its distances from two other points (the foci) is constant, or resulting when a cone is cut by an oblique plane which does not intersect the base.

Given: Ellipse equation: $49x^{2} + y^{2} = 49$

To find: Find the foci and trace the graph of the given ellipse.

Approach:

When ellipse is of the form $x^{2} /a^{2} + y^{2} /b^{2} = 1$ , coordinates of foci are of the form (0,be) when b > a and e = $\sqrt{1 - \frac{a^{2} }{b^{2} }$.

Step-by-step explanation:

Step 1 of 3:

Given ellipse is: $x^{2} + y^{2}/49 = 1$

Step 2 of 3:

Therefore, here, e =  $\sqrt{1 - \frac{1^{2} }{7^{2} }$ = $4\sqrt{3} /7$ ; Hence, (be) = $4\sqrt{3}$.

Step 3 of 3:

Therefore, coordinates of foci of the given ellipse are  $( 0, 4\sqrt{3} ) and (0, -4\sqrt{3} )$

Final Answer:

Foci are $( 0, 4\sqrt{3} ) and (0, -4\sqrt{3} )$ respectively.

Graph of given ellipse is: