JEE 2016 :: Main 2016 :: Mathematics 2016

• Main 2016 - Physics 2016
• Main 2016 - Chemistry 2016
• Main 2016 - Mathematics 2016

The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is 

Answer:

Explanation:

We have ,  $\frac{2b^{2}}{a}=8$ and 2b=ae

$\Rightarrow b^{2}=4a$ and 2b=ae

consider 2b=ae

$\Rightarrow 4b^{2}=a^{2}e^{2}$

$\Rightarrow 4a^{2}(e^{2}-1)=a^{2}e^{2}$

$\Rightarrow 4e^{2}-4=e^{2} [\therefore a\neq 0]$

$\Rightarrow 3e^{2}=4$

$\Rightarrow e=\frac{2}{\sqrt{3}}   [\because e>0]$

Let P be the point on the parabola, y²=8x, which is a minimum distance from the centre  C of the circle,  $x^{2}+(y+6)^{2}=1$ Then, the equation of the circle passing through C and having its centre at P is

Answer:

Explanation:

Centre of circle $x^{2}+(y+6)^{2}=1$ is C (0,-6)

 Let the coordinates of point $(2t^{2},4t)$

Now let D=CP

   $=\sqrt{(2t^{2})^{2}-(4t+6)^{2}}$

$\Rightarrow D=\sqrt{4t^{4}-16t^{2}+36+48t}$

squaring on both sides

$\Rightarrow D^{2}(t)=4t^{4}+16t^{2}+48t+36$

 Let $F(t)=4t^{4}+16t^{2}+48t+36$

 for minimum,

  F'(t)=0

$\Rightarrow 16t^{3}+32t+48=0$

$\Rightarrow t^{3}+2t+3=0$

$\Rightarrow (t+1)(t^{2}-t+3)=0$

 $\Rightarrow t=-1$

   Thus, coordinate of point P are (2,-4) 

Now,

   $CP= \sqrt{2^{2}+(-4+6)^{2}}=\sqrt{4+4}$

$2\sqrt{2}$

Hence the required equation of circle is 

$(x-2)^{2}+(y+4)^{2}=(2\sqrt{2}^{2})$

$x^{2}+4-4x+y^{2}+16+8y=8$

$x^{2}+y^{2}-4x+8y+12=0$

If one of the diameters of the circle, given by the equation  $x^{2}+y^{2}-4x+6y-12=0$, is a chord of a circle S, whose centre is at (-3,2), then the radius of S is

Answer:

Explanation:

Given the equation of a circle is  $x^{2}+y^{2}-4x+6y-12=0$, whose centre is (2,-3) and radius

    =$\sqrt{2^{2}+(-3)^{2}+12}=\sqrt{4+9+12}=5$

 Now , according to given information , we have the following figure

   $x^{2}+y^{2}-4x+6y-12=0$

 Clarly, $AO\perp BC$, as O is  mid-point of the  chord

Now, in ΔAOB , we have

$OA=\sqrt{(-3-2)^{2}+(2+3)^{2}}$ 

  $=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$

and OB=5

   $\therefore  AB= \sqrt{OA^{2}+OB^{2}}$

$\sqrt{50+25}=\sqrt{75}$

  = $5\sqrt{3}$

Two sides of a rhombus are along the lines, x-y+1=0 and 7x-y-5=0. If its diagonals interset at (-1,-2)  then which one of the following is a vertex of this rhombus?

Answer:

Explanation:

 As the  given lines x-y+1=0 and 7x-y-5=0 are not parallel, therefore they represent the adjacent sides of the rhombus

   On solving x-y+1=0 and 7x+y-5=0 , we get x=1 and y=2, thus, one of the vertex is A(1,2)

Let the coordinates of point C be (x,y)

 Then,  $-1=\frac{x+1}{2}$ and $-2=\frac{y+2}{2}$

  $\Rightarrow x+1=-2$ and $y=-4-2$

$\Rightarrow x=-3$ and $y=-6$

 Hence, coordinates of C =(-3,-6)

Note that vertices B and D will satisfy x-y+1=0 and 7x-y-5=0

 respectively

 Since option(c) satisfies 7x-y-5=0, therefore coordinates of vertex D is $(\frac{1}{3},-\frac{8}{3})$

 If curve y=f(x) passes through the point (1,-1) and satisfies the differential equation

y(1+xy)dx=xdy , then f(-1/2) is equal to 

Answer:

Explanation:

Given  differential equation is 

y(1+xy)dx=xdy

$\Rightarrow y dx+xy^{2}dx=x dy$

$\Rightarrow \frac{x dy -y dx}{y^{2}}=xdx$

$\Rightarrow -\frac{(y dx -x dy)}{y^{2}}=xdx$

$\Rightarrow -d(\frac{x}{y})=xdx$

 On  integrating  both sides , we get

$-\frac{x}{y}=\frac{x^{2}}{2}+C$

 It passes through(1,-1)

$1=\frac{1}{2}+C$

$\Rightarrow C=\frac{1}{2}$

Now, from (i)

$-\frac{x}{y}=\frac{x^{2}}{2}+\frac{1}{2}$

$\Rightarrow x^{2}+1=-\frac{2x}{y}$

$\Rightarrow y=-\frac{x^{2}}{2x+1}$

$f(-\frac{1}{2})=\frac{4}{5}$

• Main 2016 - Physics 2016
• Main 2016 - Chemistry 2016
• Main 2016 - Mathematics 2016