141.

The eccentricity of the hyperbola which passes through (3, 0) and (3$\sqrt{2}$, 2) is

• $\sqrt{\left(13\right)}$

• $\frac{\sqrt{13}}{3}$

• $\sqrt{\frac{13}{4}}$

• None of these

The line x - 1 = 0 is the directrix of the parabola y² - kx + 8 = 0. Then, one of the value of k is

• $\frac{1}{8}$

• 8

• 4

• $\frac{1}{4}$

The curve represented by

$\mathrm{x} = 3\left(\cos \left(\mathrm{t}\right) + \sin \left(\mathrm{t}\right)\right), \mathrm{y} = 4\left(\cos \left(\mathrm{t}\right) - \sin \left(\mathrm{t}\right)\right)$ is

• ellipse

• parabola

• hyperbola

• circle

If 5x - 12y + 10 = 0 and 12y - 5x + 16 = 0 are two tangents to a circle, then the radius of the circle is

• 1

• 2

• 4

• 6

The area of the smaller segment cut off from the circle x² + y² = 9 by x = 1 is

• $\frac{1}{2}\left(9{\sec }^{-1}\left(3 - \sqrt{8}\right)\right)$

• $\left(9{\sec }^{-1}\left(3 - \sqrt{8}\right)\right)$ sq unit

• $\left(\sqrt{8} - 9{\sec }^{-1}\left(3 \right)\right)$

• None of the above

An equilateral triangle is inscribed in the parabola y² = 4ax, whose vertex is at the vertex of the parabola. The length of its side is

• 2a$\sqrt{3}$

• 4a$\sqrt{3}$

• $6\mathrm{a}\sqrt{3}$

• 8a$\sqrt{3}$

The period of the function f(x) = $\sin \left(\sin \left(\frac{\mathrm{x}}{5}\right)\right)$ is

• $2\mathrm{\pi }$

• $\frac{2\mathrm{\pi }}{5}$

• $10\mathrm{\pi }$

• $5\mathrm{\pi }$

The foci of a hyperbola are (- 5, 18) and (10, 20) and it touches the Y-axis. The length of its transverse axis is

• 100

• $\raisebox{1ex}{$\sqrt{89}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

• $\sqrt{89}$

• $\sqrt{50}$

The locus of mid-point of the line segment joining the locus to a moving point on the parabola y² = 4ax is another parabola with directrix

• x = - a

• x = a

• x = 0

• x = a/2

The equation of the ellipse with its centre at (1, 2), one locus at (6, 2) and passing through (4, 6) is

• $\frac{{\left(\mathrm{x} - 1\right)}^2}{45} + \frac{{\displaystyle {\left(\mathrm{y} - 2\right)}^2}}{20} = 1$

• $\frac{{\left(\mathrm{x} - 1\right)}^2}{20} + \frac{{\displaystyle {\left(\mathrm{y} - 2\right)}^2}}{45} = 1$

• $\frac{{\left(\mathrm{x} + 1\right)}^2}{45} + \frac{{\displaystyle {\left(\mathrm{y} + 2\right)}^2}}{20} = 1$

• None of the above