Question

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

• A. $x^2+y^2=5$
• B. $\sqrt{5}(x^2+y^2)-3x-4y=0$
• C. $\sqrt{5}(x^2+y^2)+3x+4y=0$
• D. $x^2+y^2=25$

Answer 1

The correct option is A $x^2+y^2=5$

$\begin{array}{c}Ellipseequ:-\\ E=4x^2+9y^2=36\\ \frac{x^2}{9}+\frac{y^2}{4}=1\\ hyperbolaequ:-\\ H=4x^2-y^2=4\\ \frac{x^2}{\frac{4}{a^2}}-\frac{y^2}{4}=1\\ \left[E\&Hareorthogonal,sotheyareconfocal.............\right]\\ focusofE\&H:\\ E:s=(\pm ae,0)\\ a=3\\ e=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}\\ \\ H:s=(\pm ac,0)\\ a=\frac{2}{a}\\ e=\sqrt{1+\frac{4}{\frac{4}{a^2}}}=\sqrt{1+a^2}\\ Now,\\ \Rightarrow 3\times \frac{\sqrt{5}}{3}=\frac{2}{a}\sqrt{1+a^2}\\ \Rightarrow 5a^2=4+4a^2\\ \Rightarrow a^2=4\\ \therefore e=a=2\\ Ellipseequ:4x^2+9y^2=36----------\left(i\right)\\ Hyperbolaequ:4x^2-y^2=4--------\left(ii\right)\\ pointofintersetion:(x_1,y_1)\\ Lets,addingbothequ\left(i\right)\&\left(ii\right)\\ \underset{–}{\begin{array}{c}4x^2+9y^2=36\\ 4x^2-y^2=4\end{array}}\\ 8x^2+8y^2=40\\ \therefore x^2+y^2=5\\ So,thatthecorrectoptionisA.\end{array}$