The ellipse 4x^2 + 9y^2 = 36 and the hyperbol | vedclass

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(A) The given equations are $4x^2 + 9y^2 = 36$ (Ellipse) and $4x^2 - y^2 = 4$ (Hyperbola).Dividing the ellipse equation by $36$,we get $\frac{x^2}{9}

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$x^2 + y^2 = 5$

Vedclass · Answer

(A) The given equations are $4x^2 + 9y^2 = 36$ (Ellipse) and $4x^2 - y^2 = 4$ (Hyperbola).Dividing the ellipse equation by $36$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.Dividing the hyperbola equation by $4$,we get $\frac{x^2}{1} - \frac{y^2}{4} = 1$.Let the points of intersection be $(x_1, y_1)$.Adding the two equations: $(4x^2 + 9y^2) + k(4x^2 - y^2) = 36 + 4k$ represents a family of curves passing through the intersection points.For this to be a circle,the coefficient of $x^2$ must equal the coefficient of $y^2$ and the $xy$ term must be zero.$4 + 4k = 9 - k$ $\Rightarrow 5k = 5$ $\Rightarrow k = 1$.Substituting $k = 1$ into the equation: $(4x^2 + 9y^2) + 1(4x^2 - y^2) = 36 + 4(1)$.$8x^2 + 8y^2 = 40 \Rightarrow x^2 + y^2 = 5$.

The ellipse $4x^2 + 9y^2 = 36$ and the hyperbola $4x^2 - y^2 = 4$ have the same foci and they intersect at right angles. Then,the equation of the circle passing through the points of intersection of the two conics is:

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