The area of the region bounded by the line $y=x+2$ and the curve $x=-y^2$ is
13.5 sq units
$\frac{7}{6}$ sq units
4.5 sq units
2.5 sq units
If $x=4$ is a root of $\left|\begin{array}{cc}x & 3 \\ 1 & x-2\end{array}\right|=5$, then the other root is:
-4
3
-2
-1
The difference between the distance of any point on the hyperbola from the two foci is $\mathbf{1 6}$ and the eccentricity is $\mathbf{2}$. Then the equation of the hyperbola is
$\frac{x^2}{64}-\frac{y^2}{64}=1$
$\frac{x^2}{64}-\frac{y^2}{256}=1$
$\frac{x^2}{64}-\frac{y^2}{192}=1$
$\frac{x^2}{192}-\frac{y^2}{64}=1$
$$ \text { If } x=a(\theta-\sin \theta) \text { and } y=a(1-\cos \theta) \text {, then } \frac{\left(\mathbf{1}+\boldsymbol{y}_{\mathbf{1}}{ }^{\mathbf{2}}\right)^{\mathbf{3} / \mathbf{2}}}{\boldsymbol{y}_{\mathbf{2}}}= $$
$$ -4 a \sin \left(\frac{\theta}{2}\right) $$
$$ -\frac{1}{2 \sin ^2\left(\frac{\theta}{2}\right)} $$
$$ 4 a \operatorname{cosec}^2\left(\frac{\theta}{2}\right) $$
$$ -\frac{1}{4 a} \sin \left(\frac{\theta}{2}\right) $$
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