For a > 0, let the curves C1 : y2 = ax and C2 : x2 = 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 area bounded by the curves, C1 and C2, and the area of \Delta OQR = 1 \over 2 , then 'a' satisfies the equation :

Area Under the Curves PYQ — JEE Main 2020 · 8 Jan · Shift 1 · Q32 with Solution

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Correct answer: NONE OF THE OPTIONS IS CORRECT; THE CORRECT EQUATION IS $X^6-4X^3+2=0$ (WITH $X=\SQRT{A}$).

Find the intersection points of the curves

The curves are $C_1: y^2=ax$ and $C_2: x^2=ay$ with $a>0$ .

They intersect at the origin $O=(0,0)$ and another point $P$ .

For the non-origin intersection, from $C_2$ : $y=\frac{x^2}{a}$ Substitute into $C_1$ : $\left(\frac{x^2}{a}\right)^2=ax$ $\frac{x^4}{a^2}=ax$ $x^4=a^3x$ $x(x^3-a^3)=0$ So, for $x\neq 0$ , $x=a$ Hence, $y=\frac{a^2}{a}=a$ Thus, $P=(a,a)$

So the chord $OP$ is the line $y=x$

Coordinates of $Q$ and $R$

The line $x=b$ intersects:

the chord $OP: y=x$ at $Q=(b,b)$ ,
the $x$ -axis at $R=(b,0)$ .

Thus triangle $OQR$ has vertices $O=(0,0),\quad Q=(b,b),\quad R=(b,0)$ Its area is $[\Delta OQR]=\frac12\cdot OR\cdot RQ=\frac12\cdot b\cdot b=\frac{b^2}{2}$ Given $\frac{b^2}{2}=\frac12$ we get $b^2=1\implies b=1$ (since $b>0$ ).

Area bounded by the two curves

Between $x=0$ and $x=a$ ,

from $C_1$ : $y=\sqrt{ax}$
from $C_2$ : $y=\frac{x^2}{a}$

So total bounded area is $A=\int_0^a\left(\sqrt{ax}-\frac{x^2}{a}\right)dx$

Now,

=\sqrt a\cdot \frac{2}{3}a^{3/2}=\frac{2}{3}a^2$$ and $$\int_0^a \frac{x^2}{a}dx=\frac{1}{a}\cdot \frac{a^3}{3}=\frac{a^2}{3}$$ Therefore, $$A=\frac{2}{3}a^2-\frac{1}{3}a^2=\frac{a^2}{3}$$ Since $x=b$ bisects this area, area from $x=0$ to $x=b$ equals half of total area: $$\int_0^b\left(\sqrt{ax}-\frac{x^2}{a}\right)dx=\frac12\cdot \frac{a^2}{3}=\frac{a^2}{6}$$ --- 4. **Use $b=1$ in the bisecting-area condition** Compute: $$\int_0^b \sqrt{ax}\,dx=\sqrt a\cdot \frac{2}{3}b^{3/2}$$ $$\int_0^b \frac{x^2}{a}dx=\frac{b^3}{3a}$$ So, $$\frac{2}{3}\sqrt a\, b^{3/2}-\frac{b^3}{3a}=\frac{a^2}{6}$$ With $b=1$: $$\frac{2}{3}\sqrt a-\frac{1}{3a}=\frac{a^2}{6}$$ Multiply by $6a$: $$4a\sqrt a-2=a^3$$ Since $a\sqrt a=a^{3/2}$, $$4a^{3/2}-2=a^3$$ Rearrange: $$a^3-4a^{3/2}+2=0$$ Let $$x=\sqrt a$$ Then $a^3=x^6$ and $a^{3/2}=x^3$. Hence, $$x^6-4x^3+2=0$$ --- 5. **Match with options** We derived: $$x^6-4x^3+2=0$$ But the given options are: - A: $$x^6-12x^3+4=0$$ - B: $$x^6-12x^3-4=0$$ - C: $$x^6+6x^3-4=0$$ - D: $$x^6-6x^3+4=0$$ None matches the derived equation. --- 6. **Conclusion** The correct equation satisfied by $a$ is, after putting $x=\sqrt a$, $$\boxed{x^6-4x^3+2=0}$$ So the stored answer **A** does not agree with the mathematics. It appears there may be an error in the options or in the stored answer.

More from Area Under the Curves

View written solutionFree

Correct answer: NONE OF THE OPTIONS IS CORRECT; THE CORRECT EQUATION IS $X^6-4X^3+2=0$ (WITH $X=\SQRT{A}$).

Find the intersection points of the curves

The curves are $C_1: y^2=ax$ and $C_2: x^2=ay$ with $a>0$ .

They intersect at the origin $O=(0,0)$ and another point $P$ .

So the chord $OP$ is the line $y=x$

Coordinates of $Q$ and $R$

The line $x=b$ intersects:

the chord $OP: y=x$ at $Q=(b,b)$ ,
the $x$ -axis at $R=(b,0)$ .

Area bounded by the two curves

Between $x=0$ and $x=a$ ,

from $C_1$ : $y=\sqrt{ax}$
from $C_2$ : $y=\frac{x^2}{a}$

So total bounded area is $A=\int_0^a\left(\sqrt{ax}-\frac{x^2}{a}\right)dx$

Now,

=\sqrt a\cdot \frac{2}{3}a^{3/2}=\frac{2}{3}a^2$$ and $$\int_0^a \frac{x^2}{a}dx=\frac{1}{a}\cdot \frac{a^3}{3}=\frac{a^2}{3}$$ Therefore, $$A=\frac{2}{3}a^2-\frac{1}{3}a^2=\frac{a^2}{3}$$ Since $x=b$ bisects this area, area from $x=0$ to $x=b$ equals half of total area: $$\int_0^b\left(\sqrt{ax}-\frac{x^2}{a}\right)dx=\frac12\cdot \frac{a^2}{3}=\frac{a^2}{6}$$ --- 4. **Use $b=1$ in the bisecting-area condition** Compute: $$\int_0^b \sqrt{ax}\,dx=\sqrt a\cdot \frac{2}{3}b^{3/2}$$ $$\int_0^b \frac{x^2}{a}dx=\frac{b^3}{3a}$$ So, $$\frac{2}{3}\sqrt a\, b^{3/2}-\frac{b^3}{3a}=\frac{a^2}{6}$$ With $b=1$: $$\frac{2}{3}\sqrt a-\frac{1}{3a}=\frac{a^2}{6}$$ Multiply by $6a$: $$4a\sqrt a-2=a^3$$ Since $a\sqrt a=a^{3/2}$, $$4a^{3/2}-2=a^3$$ Rearrange: $$a^3-4a^{3/2}+2=0$$ Let $$x=\sqrt a$$ Then $a^3=x^6$ and $a^{3/2}=x^3$. Hence, $$x^6-4x^3+2=0$$ --- 5. **Match with options** We derived: $$x^6-4x^3+2=0$$ But the given options are: - A: $$x^6-12x^3+4=0$$ - B: $$x^6-12x^3-4=0$$ - C: $$x^6+6x^3-4=0$$ - D: $$x^6-6x^3+4=0$$ None matches the derived equation. --- 6. **Conclusion** The correct equation satisfied by $a$ is, after putting $x=\sqrt a$, $$\boxed{x^6-4x^3+2=0}$$ So the stored answer **A** does not agree with the mathematics. It appears there may be an error in the options or in the stored answer.

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