IIT-JEE 2011 · Paper 1 · Mathematics Q56

← IIT-JEE 2011 · Paper 1

MathematicsMultiple correct+4Question 56 of 69

Let the eccentricity of the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ be reciprocal to that of the ellipse $x^2+4y^2=4$ . If the hyperbola passes through a focus of the ellipse, then

the equation of the hyperbola is

\dfrac{x^2}3-\dfrac{y^2}2=1

a focus of the hyperbola is

(2,0)

the eccentricity of the hyperbola is

\sqrt{\dfrac53}

the equation of the hyperbola is

x^2-3y^2=3

Answer and worked solution

Official answer: B, D

Ellipse $\frac{x^2}4+y^2=1$ : $e=\frac{\sqrt3}2$ , foci $(\pm\sqrt3,0)$ .

Hyperbola. $e=\frac2{\sqrt3}$ , so $b^2=a^2(e^2-1)=\frac{a^2}3$ ; through $(\sqrt3,0)$ : $a^2=3$ , $b^2=1$ :
$\frac{x^2}3-y^2=1\iff x^2-3y^2=3\ \text{(D)},\qquad\text{foci }(\pm ae,0)=(\pm2,0)\ \text{(B)}.$

(B), (D)

Topics:Hyperbola Ellipse

← Previous question Next question →

IIT-JEE 2011 · Paper 1, Mathematics question 56