edited by
976 views
1 votes
1 votes
The area bounded by the curves $y^2$ = 9x, x - y + 2 = 0 is given by
a) 1

b) 1/2

C) 3/2

d) 5/4
edited by

2 Answers

1 votes
1 votes

In the following problem first, we have a parabola ($y^{2}= 4ax$) and the equation of a straight line given by ($y = mx + c$). We see the graph comes like this.

So , now let us equate the 2 equations:

$x + 2 = 3\sqrt{x}$

$(x + 2) ^{^{2}}= 9x$

$x^{^{2}} - 5x + 4 = 0$

By simplifying this quadratic equation we get $x = 4$ and $x = 1$ .

Similarly, substitute the values of x in the equation, we get $y = 6$  and $y = 3$.

Now by using Integration, we solve the problem

Area bounded by 2 curves is given by area bounded by the parabola - Area bounded by the St Line 

edited by
0 votes
0 votes
When d two equations are solved then the following values of y are obtained:

At y=6, x=4.

And for y=3, x=1. The area under d curve therefore varies accordingly.

Related questions

0 votes
0 votes
1 answer
1
sh!va asked Mar 8, 2017
1,736 views
Area bounded by the parabola 2y= $x^2$ and the line x = y-4 is equal to(a) 4.5(b) 9(c) 18(d) 36
1 votes
1 votes
0 answers
2
Tuhin Dutta asked Jan 25, 2018
596 views
x00.30.60.91.21.51.82.12.4f(x)00.090.360.811.442.253.244.415.76The value of the below integral computed using the continuous at x = 3?$$\int_{0}^{3} f(x) dx$$a) 8.983b) 9...
1 votes
1 votes
1 answer
3
Shubhanshu asked Sep 8, 2017
826 views
Let f(x)=x−(1/2) and A denote the area of region bounded by f(x) and the x-axis, when x varies from -1 to 1.A is nonzero and finite??
0 votes
0 votes
2 answers
4
sh!va asked Mar 10, 2017
804 views
The integrating factor of equation sec 2 y dy/dx + x tan y = x3 isa) $[e]^{x^2/2}$b) $[e]^ {-x^2/2}$c) $[e]^{x/2}$d)$ [e]^{-x/2}$