 ### JEE Mains Previous Years Questions with Solutions

4.5     (100k+ )
1

### AIEEE 2008

A parabola has the origin as its focus and the line $$x=2$$ as the directrix. Then the vertex of the parabola is at
A
$$(0,2)$$
B
$$(1,0)$$
C
$$(0,1)$$
D
$$(2,0)$$

## Explanation Vertex of a parabola is the mid point of focus and the point

where directrix meets the axis of the parabola.

Here focus is $$O\left( {0,0} \right)$$ and directrix meets the axis at $$B\left( {2,0} \right)$$

$$\therefore$$ Vertex of the parabola is $$(1,0)$$
2

### AIEEE 2008

A focus of an ellipse is at the origin. The directrix is the line $$x=4$$ and the eccentricity is $${{1 \over 2}}$$. Then the length of the semi-major axis is
A
$${{8 \over 3}}$$
B
$${{2 \over 3}}$$
C
$${{4 \over 3}}$$
D
$${{5 \over 3}}$$

## Explanation Perpendicular distance of directrix from focus

$$= {a \over e} - ae = 4$$

$$\Rightarrow a\left( {2 - {1 \over 2}} \right) = 4$$

$$\Rightarrow a = {8 \over 3}$$

$$\therefore$$ Semi major axis $$=8/3$$
3

### AIEEE 2007

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
4

### AIEEE 2007

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 $$\left( { - 2,0} \right)$$

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12