Get Answers to all your Questions

header-bg qa

For a>0, let the curves C_{1}:y^{2}=ax and C_{2}:x^{2}=ay intersect at origin O and a point P. Let the line x=b(0<b<a) intersect the chord OP and the x-axis at points Q and R,respectively. If the line x=b bisects the are bounded by the curves, C_{1} and C_{2'} and the area of \Delta OQR=\frac{1}{2}, then 'a' satisfies the equation :
 
Option: 1 x^{6}-12x^{3}+4=0
Option: 2 x^{6}-12x^{3}-4=0
Option: 3 x^{6}+6x^{3}-4=0
Option: 4 x^{6}-6x^{3}+4=0
 

Answers (1)

best_answer

 

 

Parabola -

Parabola

A parabola is the set of all points (x, y) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix in the plane.

Standard equation of a parabola

Let focus of parabola is S(a, 0) and directrix be x + a = 0, and axis as x-axis 

P(x, y) is any point on the parabola.

Now, from the definition of the parabola, 
\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;SP=PM}\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;SP^2=PM^2}\\\mathrm{\Rightarrow \;\;\;\;\;(x-a)^{2}+(y-0)^{2}=(x+a)^{2}}\\\mathrm{\Rightarrow \;\;\;\;\;y^2=4ax}

which is the required equation of a standard parabola

-

 

 

Area Bounded by Curves When Intersects at More Than One Point -

Area bounded by the curves  y = f(x),  y = g(x)  and  intersect each other in the interval [a, b]

First find the point of intersection of these curves  y = f(x) and  y = g(x) , let the point of intersection be x = c

Area of the shaded region  

=\int_{a}^{c}\{f(x)-g(x)\} d x+\int_{c}^{b}\{g(x)-f(x)\} d x

 

When two curves intersects more than one point

rea bounded by the curves  y=f(x),  y=g(x)  and  intersect each other at three points at  x = a, x = b amd x = c.

To find the point of intersection, solve f(x) = g(x).

For x ∈ (a, c), f(x) > g(x) and for x ∈ (c, b), g(x) > f(x).

Area bounded by curves,

\\\mathrm{A=} \int_{a}^{b}\left |f(x)-g(x) \right |dx\\\\\mathrm{\;\;\;\;=} \int_{a}^{c}\left ( f(x)-g(x) \right )dx+\int_{c}^{b}\left ( g(x)-f(x) \right )dx  

 

-

 

 

\begin{array}{l}{\frac{1}{2} (b \times b)=\frac{1}{2}} \\ {b=1}\end{array}

\int_{0}^{1}\left(\sqrt{a} \sqrt{x}-\frac{x^{2}}{a}\right) d x=\frac{a^2}{6} \text{ by property of parabola}

By solving above you will get

a^{6}-12 a^{3}+4=0

Correct option (1)

Posted by

Kuldeep Maurya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE