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

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

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

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

• None of these

The centre of circle lies on x + y = 3 and touching the lines x = 3, y = 3 then find diameter of circle

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

The circle passing through the intersection of the circles, x² + y² – 6x = 0 and x² + y² – 4y = 0, having its centre on the line, 2x – 3y + 12 = 0, also passes through the point :

• $\left(1, - 3\right)$

• $\left(- 1, 3\right)$

• ( - 3, 6)

• ( - 3, 1)

$$\begin{gathered} \mathrm{Let} \mathrm{x} = 4 \mathrm{be} \mathrm{a} \mathrm{directrix} \mathrm{to} \mathrm{an} \mathrm{ellipse} \mathrm{whose} \mathrm{centre} \mathrm{is} \mathrm{at} \mathrm{the} \mathrm{origin} \mathrm{and} \mathrm{its} \mathrm{eccentricity} \mathrm{is} 21. \\ \mathrm{If} \mathrm{P}(1, \mathrm{\beta }), \mathrm{\beta } > 0 \mathrm{is} \mathrm{a} \mathrm{point} \mathrm{on} \mathrm{this} \mathrm{ellipse}, \mathrm{then} \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{normal} \mathrm{to} \mathrm{it} \mathrm{at} \mathrm{P} \mathrm{is} \end{gathered}$$

• 7x - 4y = 1

• 4x - 2y = 1

• 8x - 2y = 5

• 4x - 3y = 2

$$\begin{gathered} \mathrm{Let} \mathrm{PQ} \mathrm{be} \mathrm{a} \mathrm{diameter} \mathrm{of} \mathrm{the} \mathrm{circle} {\mathrm{x}}^2 + {\mathrm{y}}^2 = 9. \mathrm{If} \mathrm{\alpha } \mathrm{and} \mathrm{\beta } \mathrm{are} \mathrm{the} \mathrm{lengths} \mathrm{of} \mathrm{the} \mathrm{perpendiculars} \mathrm{from} \\ \mathrm{P} \mathrm{and} \mathrm{Q} \mathrm{on} \mathrm{the} \mathrm{straight} \mathrm{line}, \mathrm{x} + \mathrm{y} = 2 \mathrm{respectively}, \mathrm{then} \mathrm{the} \mathrm{maximum} \mathrm{value} \mathrm{of} \mathrm{\alpha \beta } \mathrm{is} ..... \end{gathered}$$

If the common tangent to the parabolas, y² = 4x and x² = 4y also touches the circle, x² + y² = c², then c is equal to :

• $\frac{1}{\sqrt{2}}$

• $\frac{1}{2\sqrt{2}}$

• $\frac{1}{2}$

• $\frac{1}{4}$