IIT-JEE 2007 · Paper 1 · Mathematics Q50

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MathematicsSingle correct+3 / -1Question 50 of 66

A hyperbola, having the transverse axis of length $2\sin\theta$ , is confocal with the ellipse $3x^2+4y^2=12$ . Then its equation is

x^2\operatorname{cosec}^2\theta-y^2\sec^2\theta=1

x^2\sec^2\theta-y^2\operatorname{cosec}^2\theta=1

x^2\sin^2\theta-y^2\cos^2\theta=1

x^2\cos^2\theta-y^2\sin^2\theta=1

Answer and worked solution

Official answer: A

Ellipse. $\dfrac{x^2}4+\dfrac{y^2}3=1$ : $e^2=1-\frac34=\frac14$ , foci $(\pm1,0)$ .

Hyperbola. $a=\sin\theta$ and $ae=1$ , so $b^2=a^2e^2-a^2=1-\sin^2\theta=\cos^2\theta$ :
$\frac{x^2}{\sin^2\theta}-\frac{y^2}{\cos^2\theta}=1,\quad\text{i.e. }x^2\operatorname{cosec}^2\theta-y^2\sec^2\theta=1.$

(A)

Topics:Hyperbola Ellipse

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IIT-JEE 2007 · Paper 1, Mathematics question 50