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1

### AIEEE 2007

MCQ (Single Correct Answer)
The normal to a curve at $$P(x,y)$$ meets the $$x$$-axis at $$G$$. If the distance of $$G$$ from the origin is twice the abscissa of $$P$$, then the curve is a
A
circle
B
hyperbola
C
ellipse
D
parabola

## Explanation

Equation of normal at $$P\left( {x,y} \right)$$ is $$Y - y = - {{dx} \over {dy}}\left( {x - x} \right)$$

Coordinate of $$G$$ at $$X$$ axis is $$\left( {X,0} \right)$$ (let)

$$\therefore$$ $$0 - y = - {{dx} \over {dy}}\left( {X - x} \right) \Rightarrow y{{dy} \over {dx}} = X - x$$

$$\Rightarrow X = x + y{{dy} \over {dx}}$$ $$\therefore$$ Co-ordinate of $$G\left( {x + y{{dy} \over {dx}},0} \right)$$

Given distance of $$G$$ from origin $$=$$ twice of the abscissa of $$P.$$

as distance cannot be $$-ve,$$ therefore abscissa $$x$$ should be $$+ve$$

$$\therefore$$ $$x + y{{dy} \over {dx}} = 2x \Rightarrow y{{dy} \over {dx}} = x \Rightarrow ydx = xdx$$

On Integrating $$\Rightarrow {{{y^2}} \over 2} = {{{x^2}} \over 2} + {c_1} \Rightarrow {x^2} - {y^2} = - 2{c_1}$$

$$\therefore$$ the curve is a hyperbola
2

### AIEEE 2007

MCQ (Single Correct Answer)
The equation of a tangent to the parabola $${y^2} = 8x$$ is $$y=x+2$$. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is
A
$$(2,4)$$
B
$$(-2,0)$$
C
$$(-1,1)$$
D
$$(0,2)$$

## Explanation

Parabola $${y^2} = 8x$$

We know that the locus of point of intersection of two perpendicular tangents to a

parabola is its directrix. Point must be on the directrix of parabola

as equation of directrix $$x + 2 = 0 \Rightarrow x = - 2$$

Hence the point is $$\left( { - 2,0} \right)$$
3

### AIEEE 2007

MCQ (Single Correct Answer)
For the Hyperbola $${{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1$$ , which of the following remains constant when $$\alpha$$ varies$$=$$?
A
abscissae of vertices
B
abscissaeof foci
C
eccentricity
D
directrix.

## Explanation

Given, equation of hyperbola is $${{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1$$

We know that the equation of hyperbola is

$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ Here, $${a^2} = {\cos ^2}\alpha$$ and $${b^2} = {\sin ^2}\alpha$$

We know that, $${b^2} = {a^2}\left( {{e^2} - 1} \right)$$

$$\Rightarrow {\sin ^2}\alpha = {\cos ^2}\alpha \left( {{e^2} - 1} \right)$$

$$\Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = {\cos ^2}\alpha .{e^2}$$

$$\Rightarrow {e^2} = 1 + {\tan ^2}\alpha = {\sec ^2}\alpha \Rightarrow e = \sec \,\alpha$$

$$\therefore$$ $$ae = \cos \alpha \,\,.\,\,{1 \over {\cos \alpha }} = 1$$

Co-ordinate of foci are $$\left( { \pm \alpha e,0} \right)\,\,$$ i.e. $$\left( { \pm 1,0} \right)$$

Hence, abscissae of foci remain constant when $$\alpha$$ varies.
4

### AIEEE 2006

MCQ (Single Correct Answer)
Angle between the tangents to the curve $$y = {x^2} - 5x + 6$$ at the points $$(2,0)$$ and $$(3,0)$$ is
A
$$\pi$$
B
$${\pi \over 2}$$
C
$${\pi \over 6}$$
D
$${\pi \over 4}$$

## Explanation

$${{dy} \over {dx}} = 2x - 5$$

$$\therefore$$ $${m_1} = {\left( {2x - 5} \right)_{\left( {2,0} \right)}} = - 1,$$

$${m_2} = {\left( {2x - 5} \right)_{\left( {3,0} \right)}} = 1 \Rightarrow {m_1}{m_2} = - 1$$

i.e. the tangents are perpendicular to each other.

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