A hyperbola passing through a focus of the  ellipse$\frac{{\mathrm{x}}^2}{169} + \frac{{\mathrm{y}}^2}{25} = 1$. Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of eccentricities is 1. Then, the equation of the hyperbola is

• $\frac{{\mathrm{x}}^2}{144} - \frac{{\mathrm{y}}^2}{9} = 1$

• $\frac{{\mathrm{x}}^2}{169} - \frac{{\mathrm{y}}^2}{25} = 1$

• $\frac{{\mathrm{x}}^2}{144} - \frac{{\mathrm{y}}^2}{25} = 1$

• $\frac{{\mathrm{x}}^2}{125} - \frac{{\mathrm{y}}^2}{9} = 1$

$$\begin{gathered} \mathrm{If} \mathrm{the} \mathrm{curves} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2} + \frac{{\mathrm{y}}^2}{{\mathrm{b}}^2} = 1 \mathrm{and} \frac{{\mathrm{x}}^2}{25} + \frac{{\mathrm{y}}^2}{16} = 1 \\ \mathrm{cut} \mathrm{each} \mathrm{other} \mathrm{orthogonally}, \mathrm{then} {\mathrm{a}}^2 - {\mathrm{b}}^2 = ? \end{gathered}$$

• 9

• 400

• 75

• 41

Te area (m sq units) of the region bounded by x = -1, x = 2, y = x² + 1 and y = 2x - 2 is

• 10

• 7

• 8

• 9

The sum of the minimum and maximum distance of the point (4, - 3) to the circle x² + y² + 4x - 10y - 7 = 0, is

• 10

• 12

• 16

• 20

The locus of centres of the circles, which cut the circles x² + y² + 4x - 6y + 9 and x² + y² - 5x + 4y + 2 = 0 orthogonally, is

• 3x + 4y - 5 = 0

• 9x - 10y + 7 = 0

• 9x + 10y - 7 = 0

• 9x - 10y + 11 = 0

If x - y + 1 = 0 meets the circle x² + y² + y - 1 = 0 at A and B, then the equation of the circle with AB as diameter is

• 2(x² + y²) + 3x - y + 1 = 0

• 2(x² + y²) + 3x - y + 2 = 0

• 2(x² + y²) + 3x - y + 3 = 0

• x² + y² + 3x - y + 4 = 0

An equilateral triangle is inscribed in the parabola y² = Bx, with one of its vertices is the vertex of the parabola. Then, length of the side of that triangle is

• $24\sqrt{3} \mathrm{units}$

• $16\sqrt{3}$ units

• $8\sqrt{3} \mathrm{units}$

• $4\sqrt{3} \mathrm{units}$

The point (3, 4) is the focus and 2x - 3y + 5 = 0 is the directrix of a parabola. Its latus rectum is

• $\frac{2}{\sqrt{13}}$

• $\frac{4}{\sqrt{13}}$

• $\frac{1}{\sqrt{13}}$

• $\frac{3}{\sqrt{13}}$

The radius of the circle passing through the foci of the ellipse $\frac{{\mathrm{x}}^2}{16} + \frac{{\mathrm{y}}^2}{9} = 1$ and having its centre at (0, 3) is

• 6

• 4

• 3

• 2

The equation of the circle passing through (2, 0) and (0, 4) and having the minimum radius, is

• x² + y² = 20

• x² + y² - 2x - 4y = 0

• x² + y² = 4

• x² + y² = 16