Conic Section

A circle having centre at the origin passes through the three vertices of an equilateral triangle the length of its median being 9 units. Then the equation of that circle is

• ${\mathrm{x}}^2 + {\mathrm{y}}^2 = 9$

• ${\mathrm{x}}^2 + {\mathrm{y}}^2 = 18$

• ${\mathrm{x}}^2 + {\mathrm{y}}^2 = 36$

• ${\mathrm{x}}^2 + {\mathrm{y}}^2 = 81$

A line parallel to the straight line 2x – y = 0 is tangent to the hyperbola $\frac{{\mathrm{x}}^2}{4} - \frac{{\mathrm{y}}^2}{2} = 1$  at the point (x₁, y₁). Then ${{\mathrm{x}}_1}^2 + 5{{\mathrm{y}}_1}^2$ is equal to :

• 10

• 5

• 8

• 6

The number of integral values of k for which the line, 3x + 4y = k intersects the circle, x² + y² – 2x – 4y + 4 = 0 at two distinct points is .... 

For some $\mathrm{\theta } \in \left(0, \frac{\mathrm{\pi }}{2}\right)$, if the eccentricity of the hyperbola, ${\mathrm{x}}^2 - {\mathrm{y}}^2{\sec }^2\left(\mathrm{\theta }\right) = 10$ is 5 times the eccentricity of the ellipse , ${\mathrm{x}}^2{\sec }^2\left(\mathrm{\theta }\right) + {\mathrm{y}}^2 = 5$, then the length of the latus rectum of the ellipse, is

• $\frac{2\sqrt{5}}{3}$

• $2\sqrt{6}$

• $\frac{4\sqrt{5}}{3}$

• $\sqrt{30}$

The area (in sq. units) of an equilateral triangle inscribed in the parabola y² = 8x, with one of its vertices on the vertex of this parabola is

• $64\sqrt{3}$

• $192\sqrt{3}$

• $128\sqrt{3}$

• $256\sqrt{3}$

$$\begin{gathered} \mathrm{Let} \frac{{\mathrm{x}}^2}{4} + \frac{{\mathrm{y}}^2}{3} = 1 \mathrm{is} \mathrm{an} \mathrm{ellipse} \mathrm{and} \mathrm{a} \mathrm{hyperbola} \mathrm{which} \mathrm{is} \mathrm{confocal} \mathrm{with} \mathrm{ellipse} \mathrm{such} \mathrm{that} \\ \mathrm{its} \mathrm{transverse} \mathrm{axis} \mathrm{is} \sqrt{2} \mathrm{then} \mathrm{which} \mathrm{of} \mathrm{following} \mathrm{point} \mathrm{does} \mathrm{not} \mathrm{lie} \mathrm{on} \mathrm{hyperbola} \end{gathered}$$

• $\left(1, - \frac{1}{\sqrt{2}}\right)$

• $\left(-\sqrt{\frac{3}{2}}, 1\right)$

• $\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}\right)$

• None of these

Let e₁ and e₂ be the eccentricities of the ellipse, $\frac{{\mathrm{x}}^2}{25} + \frac{{\displaystyle {\mathrm{y}}^2}}{{{\displaystyle \mathrm{b}}}^2} = 1\left(\mathrm{b} < 5\right) \mathrm{and} \mathrm{the} \mathrm{hyperbola}, \frac{{\mathrm{x}}^2}{16} - \frac{{\mathrm{y}}^2}{{\mathrm{b}}^2} = 1$respectively satisfying e₁e₂ = 1. If $\mathrm{\alpha } \mathrm{and} \mathrm{\beta }$ are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair $\left(\mathrm{\alpha }, \mathrm{\beta }\right)$ is equal to:

• (8, 10)

• $\left(\frac{20}{3}, 12\right)$

• (8, 12)

• $\left(\frac{24}{5}, 10\right)$

If the tangent to the curve, y = e^x at a point (c, e^c) and the normal to the parabola, y² = 4x at the point (1, 2) intersect at the same point on the x-axis, then the value of c is.....

Let P(3, 3) be a point on the hyperbola,$\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2} - \frac{{\mathrm{y}}^2}{{\mathrm{b}}^2} = 1$ . If the normal to it at P intersects the x-axis at (9,0) and e is its eccentricity, then the ordered pair (a², e²) is equal to :

• $\left(\frac{9}{2}, 3\right)$

• $\left(\frac{3}{2}, 2\right)$

• $\left(9, 3\right)$

• $\left(\frac{9}{2}, 2\right)$

Let $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2} + \frac{{\mathrm{y}}^2}{{\mathrm{b}}^2} = 1 \left(\mathrm{a} > \mathrm{b}\right)$ be agiven ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function $\mathrm{\phi }\left(\mathrm{t}\right) = \frac{5}{12} + \mathrm{t} - {\mathrm{t}}^2$ then a² + b² is equal to

• 145

• 126

• 116

• 135