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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
Let $$O$$ be the vertex and $$Q$$ be any point on the parabola, $${{x^2} = 8y}$$. If the point $$P$$ divides the line segment $$OQ$$ internally in the ratio $$1:3$$, then locus of $$P$$ is :
A
$${y^2} = 2x$$
B
$${{x^2} = 2y}$$
C
$${{x^2} = y}$$
D
$${y^2} = x$$

Explanation

Let the coordinates of Q and P be (x1, y1) and (h, k) respectively.

$$\because$$ Q lies on x2 = 8y,

$$\therefore$$ x$$_1^2$$ = 8y ....... (1)

Again, P divides OQ internally in the ratio 1 : 3.

$$\therefore$$ $$h = {{{x_1} + 0} \over 4} = {{{x_1}} \over 4}$$ or x1 = 4h and

$$k = {{{y_1} + 0} \over 4} = {{{y_1}} \over 4}$$ or y1 = 4k

Now putting x1 and y1 in (1) we get,

16h2 = 32k or, h2 = 2k

$$\therefore$$ the locus of P is given by, x2 = 2y.

2

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
The normal to the curve, $${x^2} + 2xy - 3{y^2} = 0$$, at $$(1,1)$$
A
meets the curve again in the third quadrant.
B
meets the curve again in the fourth quadrant.
C
does not meet the curve again.
D
meets the curve again in the second quadrant.

Explanation

Given curve is

$${x^2} + 2xy - 3{y^2} = 0$$

Difference $$w.r.t.x,$$

$$2x + 2x{{dy} \over {dx}} + 2y - 6y{{dy} \over {dx}} = 0$$

$${\left( {{{dy} \over {dx}}} \right)_{\left( {1,1} \right)}} = 1$$

Equation of normal at $$(1,1)$$ is

$$y=2-x$$

Solving eq. $$(1)$$ and $$(2),$$ we get $$x=1,3$$

Point of intersection $$\left( {1,1} \right),\left( {3, - 1} \right)$$

Normal cuts the curve again in $$4$$th quadrant.
3

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
The slope of the line touching both the parabolas $${y^2} = 4x$$ and $${x^2} = - 32y$$ is
A
$${{1 \over 8}}$$
B
$${{2 \over 3}}$$
C
$${{1 \over 2}}$$
D
$${{3 \over 2}}$$

Explanation

Let tangent to $${y^2} = 4x$$ be $$y = mx + {1 \over m}$$

Since this is also tangent to $${x^2} = - 32y$$

$$\therefore$$ $${x^2} = - 32\left( {mx + {1 \over m}} \right)$$

$$ \Rightarrow {x^2} + 32mx + {{32} \over m} = 0$$

Now, $$D=0$$

$${\left( {32} \right)^2} - 4\left( {{{32} \over m}} \right) = 0$$

$$ \Rightarrow {m^3} = {4 \over {32}} \Rightarrow m = {1 \over 2}$$
4

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
The locus of the foot of perpendicular drawn from the centre of the ellipse $${x^2} + 3{y^2} = 6$$ on any tangent to it is
A
$$\left( {{x^2} + {y^2}} \right) ^2 = 6{x^2} + 2{y^2}$$
B
$$\left( {{x^2} + {y^2}} \right) ^2 = 6{x^2} - 2{y^2}$$
C
$$\left( {{x^2} - {y^2}} \right) ^2 = 6{x^2} + 2{y^2}$$
D
$$\left( {{x^2} - {y^2}} \right) ^2 = 6{x^2} - 2{y^2}$$

Explanation

Given $$e{q^n}$$ of ellipse can be written as

$${{{x^2}} \over 6} + {{{y^2}} \over 2} = 1 \Rightarrow {a^2} = 6,{b^2} = 2$$

Now, equation of any variable tangent is

$$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} ....\left( i \right)$$

where $$m$$ is slope of the tangent

So, equation of perpendicular line drawn-

from center to tangent is

$$y = {{ - x} \over m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Eliminating $$m,$$ we get

$$\left( {{x^4} + {y^4} + 2{x^2}{y^2}} \right) = {a^2}{x^2} + {b^2}{y^2}$$

$$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {a^2}{x^2} + {b^2}{y^2}$$

$$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}$$

Questions Asked from Conic Sections

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