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Conic Section

Multiple Choice Questions

181.
The equation of the common tangents to the two hyperbolas $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 and \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1, are$

$y = \pm x \pm \sqrt{b^{2} - a^{2}}$

$y = \pm x \pm \sqrt{a^{2} - b^{2}}$

$y = \pm x \pm \sqrt{a^{2} + b^{2}}$

$y = \pm x \pm (a^{2} - b^{2})$

$y = \pm x \pm \sqrt{a^{2} - b^{2}}$

We have the hyperbolas

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ ...(i)

and $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$ ...(ii)

Any tangent to the hyperbola Eq. (i),

y = mx + c

where c = $\pm \sqrt{a^{2} m^{2} - b^{2}}$ ...(iii)

But this tangent touches the parabola Eq. (ii), also

$∴ \frac{{(mx + c)}^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \Rightarrow b^{2} (m^{2} x^{2} + c^{2} + 2 cmx) - a^{2} x^{2} = a^{2} b^{2} \Rightarrow (b^{2} m^{2} - a^{2}) x^{2} + 2 {mcxb}^{2} + b^{2} c^{2} - a^{2} b^{2} = 0 \Rightarrow (b^{2} m^{2} - a^{2}) x^{2} + 2 {mcb}^{2} x + b^{2} (c^{2} - a^{2}) = 0 For the tangency, it should have equal roots {(2 {mcb}^{2})}^{2} = 4 (b^{2} m^{2} - a^{2}) . b^{2} (c^{2} - a^{2})$

$\Rightarrow 4 m^{2} c^{2} b^{4} = 4 b^{2} (b^{2} m^{2} c^{2} - b^{2} m^{2} a^{2} - a^{2} c^{2} + a^{4}) \Rightarrow m^{2} c^{2} b^{2} = b^{2} m^{2} c^{2} - b^{2} m^{2} a^{2} - a^{2} c^{2} + a^{4} \Rightarrow a^{2} c^{2} = a^{4} - b^{2} m^{2} a^{2} \Rightarrow c^{2} = a^{2} - b^{2} m^{2} \Rightarrow a^{2} m^{2} - b^{2} = a^{2} - b^{2} m^{2} [∵ using Eq . (iii)] \Rightarrow (a^{2} + b^{2}) m^{2} = a^{2} + b^{2} \Rightarrow m^{2} = 1 \Rightarrow m = \pm 1$

Hence, the equation of common tangent are

y = $\pm x + \pm \sqrt{a^{2} - b^{2}}$

182.

The equation of sphere concentric with the sphere x² + y² + z² - 4x - 6y - 8z - 5 = 0 and which passes through the origin, is

x² + y² + z² - 4x - 6y - 8z = 0
x² + y² + z² - 6y - 8z = 0
x² + y² + z² = 0
x² + y² + z² - 4x - 6y - 8z - 6 = 0

183.

If two circles 2x² + 2y² - 3x + 6y + k = 0 and x² + y² - 4x + 10y + 16 = 0 cut orthagobally then, value of k is

184.

The locus of z satisfying the inequality $|\frac{z + 2 i}{2 z + i}| < 1$ , where z = x + iy, is

x² + y² < 1
x² - y² < 1
x² + y² > 1
2x² + 3y² < 1

185.

The area (in square unit) of the circle which touches the lines 4x + 3y = 15 and 4x + 3y = 5 is

4 $π$
$3 π$
$2 π$
$π$

186.

The equations of the circle which pass through the origin and makes intercepts of lengths 4 and 8 on the x and y -axes respectively are

x² + y² ± 4x ± 8y = 0
x² + y² ± 2x ± 4y = 0
x² + y² ± 8x ± 16y = 0
x² + y² ± x ± y = 0

187.

The point (3 -4) lies on both the circles x² + y² - 2x + 8y + 13 = 0 and x² + y² - 4x + 6y + 11 = 0. Then, the angle between the circles is

$60 °$
$\tan^{- 1} (\frac{1}{2})$
$\tan^{- 1} (\frac{3}{5})$
135 $°$

188.

The equation of the circle which passes through the origin and cuts orthogonally each of the circles x²+ y² - 6x + 8 = 0 and x² + y² - 2x - 2y = 7 is

3x² + 3y² - 8x - 13y = 0
3x² + 3y² - 8x + 29y = 0
3x² + 3y² + 8x + 29y = 0
3x² + 3y² - 8x - 29y = 0

189.

The number of normals drawn to the parabola y² = 4x from the point (1, 0) is

190.

If the circle x² + y² = a intersects the hyperbola xy = c²in four points (x_i, y_i), for i = 1, 2, 3 and 4, then y₁ + y₂ + y₃ + y₄ equals

0
c
a
c⁴

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Multiple Choice Questions

181. The equation of the common tangents to the two hyperbolas x2a2 - y2b2 = 1 and y2a2 - x2b2 = 1, are y = ± x ± b2 - a2 y = ± x ± a2 - b2 y = ± x ± a2 + b2 y = ± x ± a2 - b2