If a normal chord at a point t on the parabola y² = 4ax subtends a right angle at the vertex, then t equals to

• 1

• $\sqrt{2}$

• 2

• $\sqrt{3}$

The slopes of the focal chords of the parabola y² = 32x, which are tangents to the circle x² + y² = 4, are

• $\frac{1}{2}, - \frac{1}{2}$

• $\frac{1}{\sqrt{3}}, \frac{- 1}{\sqrt{3}}$

• $\frac{1}{\sqrt{15}}, \frac{- 1}{\sqrt{15}}$

• $\frac{2}{\sqrt{5}}, \frac{- 2}{\sqrt{5}}$

If tangents are drawn from any point on the circle x² + y² = 25 to the ellipse $\frac{{\mathrm{x}}^2}{16} + \frac{{\mathrm{y}}^2}{9} = 1$,  then the angle between the tangents is

• $\frac{2\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{2}$

An ellipse passing through has $\left(4\sqrt{2}, 2\sqrt{6}\right)$ foci at (- 4, 0) and (4, 0). Then, its eccentricity

• $\sqrt{2}$

• $\frac{1}{2}$

• $\frac{1}{\sqrt{2}}$

• $\frac{1}{\sqrt{3}}$

If the line joining A(1, 3, 4) and B is divided by the point    (- 2, 3, 5) in the ratio 1 : 3, then B is

• (- 11, 3 , 8)

• (- 11, 3 , - 8)

• (8, 12 , 20)

• (13, 6 , - 13)

If the direction cosines of two lines are given by l + m + n = 0 and l² - 5m² + n² = 0, then the angle between them is

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{6}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{3}$

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

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

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

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

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

If A(3, 4, 5), B(4, 6, 3), C( - 1, 2, 4) and D(1, 0, 5) are such that the angle between the lines DC and AB is $\left(\mathrm{\theta }\right)$ , then cos$\left(\mathrm{\theta }\right)$ is equal to

• $\frac{7}{9}$

• $\frac{2}{9}$

• $\frac{4}{9}$

• $\frac{5}{9}$

19.

$$\begin{gathered} \mathrm{If} \mathrm{f} :\hspace{0.17em}\left[- 2, 2\right] \to \mathrm{R} \mathrm{is} \mathrm{defined} \mathrm{by} \\ \mathrm{f}\left(\mathrm{x}\right) = \left\{\frac{\sqrt{1 + \mathrm{cx}} - \sqrt{1 - \mathrm{cx}}}{\mathrm{x}} \mathrm{for} - 2 \le \mathrm{x} \le 0 \\ \frac{\mathrm{x} + 3}{\mathrm{x} + 1} \mathrm{for} 0 \le \mathrm{x} \le 2\right\} \\ \mathrm{continuous} \mathrm{on} \left[- 2, 2\right], \mathrm{then} \mathrm{c} \mathrm{is} \mathrm{equal} \mathrm{to} \end{gathered}$$

• $\frac{2}{\sqrt{3}}$

• 3

• $\frac{3}{2}$

• $\frac{3}{\sqrt{2}}$

$\mathrm{If} \mathrm{f}\left(\mathrm{x}\right) = {\mathrm{xtan}}^{-1}\left(\mathrm{x}\right), \mathrm{then} \underset{\mathrm{x}\to 1}{\lim }\frac{\mathrm{f}\left(\mathrm{x}\right) - \mathrm{f}\left(1\right)}{\mathrm{x} - 1} = ?$

• $\frac{\mathrm{\pi } +\hspace{0.17em}3}{4}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi } + 1}{4}$

• $\frac{\mathrm{\pi } +\hspace{0.17em}2}{4}$