The line AB cuts off equal intercepts 2a from the axes. From any point P on the line AB perpendiculars PR and PS are drawn on the axes. Locus of mid-point of RS is

• x + y = a

• x² + y² = 4a²

• x² - y² = 2a²

22.

X + 8y - 22 = 0, 5x + 2y -34 = 0, 2x - 3y + 13= 0 are the three sides of a triangle. The area of the triangle is

• 36 sq units

• 19 sq units

• 42 sq units

• 72 sq units

The line through the points (a, b) and (- a,- b) passes through the point

• (1, 1)

• (3a, - 2b)

• (a², ab)

• (a, b)

The locus of the point of intersection of the straight  lines $\frac{\mathrm{x}}{\mathrm{a}} + \frac{\mathrm{y}}{\mathrm{b}} = \mathrm{K} \mathrm{and} \frac{\mathrm{x}}{\mathrm{a}} - \frac{\mathrm{y}}{\mathrm{b}} = \frac{1}{\mathrm{K}}$ where K is a non-zero real variable, is given by

• a straight line

• an ellipse

• a parabola

• a hyperbola

The equation of a line parallel to the line 3x + 4y= 0 and touching the circle x² + y² = 9 in the first quadrant, is

• 3x +4y = 15

• 3x +4y = 45

• 3x +4y = 9

• 3x +4y = 27

A line passing through the point of intersection of x + y = 4 and x - y = 2 makes an angle ${\tan }^{-1}\left(\frac{3}{4}\right)$ with the x-axis. It intersects the parabola y² = 4(x-3) at points $\left({\mathrm{x}}_1 , {\mathrm{y}}_1\right) \mathrm{and} \left({\mathrm{x}}_2, {\mathrm{y}}_2\right)$ respectively. Then, $\left|{\mathrm{x}}_1 - {\mathrm{x}}_2\right|$

• $\frac{16}{9}$

• $\frac{32}{9}$

• $\frac{40}{9}$

• $\frac{80}{9}$

The equation of auxiliary circle of the ellipse 16x² + 25y² + 32x - 100y = 284 is

• x² + y² + 2x - 4y - 20 = 0

• x² + y² + 2x - 4y = 0

• (x + 1)² + (y - 2)² = 400

• (x + 1)² + (y - 2)² = 225

If PQ is a double ordinate of the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2} - \frac{{\mathrm{y}}^2}{{\mathrm{b}}^2} = 1 \mathrm{such} \mathrm{that} ∆\mathrm{OPQ}$ is equilateral. O being the centre. Then, the eccentricity e satisfies 

• $1 < \mathrm{e} < \frac{2}{\sqrt{3}}$

• e = $\frac{2}{\sqrt{2}}$

• e = $\frac{\sqrt{3}}{2}$

• $\mathrm{e} > \frac{2}{\sqrt{3}}$

If the vertex of the conic y - 4y = 4x - 4a always lies between the straight lines x + y = 3 and 2x + 2y - 1 = 0, then

• 2 < a < 4

• $- \frac{1}{2} < \mathrm{a} < 2$

• 0 < a < 2

• $- \frac{1}{2} < \mathrm{a} < \frac{3}{2}$

$\underset{\mathrm{x}\to 1}{\lim }{\left(\frac{1 + \mathrm{x}}{2 + \mathrm{x}}\right)}^{\frac{\left(1 - \sqrt{\mathrm{x}}\right)}{\left(1 - \mathrm{x}\right)}}$ is equal to

• 1

• does not exist

• $\sqrt{\frac{2}{3}}$

• $\ln \left(2\right)$