For real numbers a, b (a > b > 0), let Area \ (x,y):x^2 + y^2 \le a^2\,and\,x^2 \over a^2 + y^2 \over b^2 \ge 1 \ = 30\pi and Area \ (x,y):x^2 + y^2 \le b^2\,and\,x^2 \over a^2 + y^2 \over b^2 \le 1 \ = 18\pi Then, the value of (a - b)2 is equal to \underline\hspace2cm .

Area Under the Curves PYQ — JEE Main 2022 · 29 Jun · Shift 2 · Q38 with Solution

View written solutionFree

Correct answer: 39-3\SQRT{69}

Interpret the two regions

We have:

Circle $C_a: x^2+y^2\le a^2$
Circle $C_b: x^2+y^2\le b^2$
Ellipse $E: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\le 1$

with $a>b>0$ .

We are given: $\text{Area}\big(C_a \cap E^c\big)=30\pi$ and $\text{Area}\big(C_b \cap E\big)=18\pi.$

We need $(a-b)^2$ .

Compare the circle and ellipse

Write both in polar form: $x=r\cos\theta,\qquad y=r\sin\theta.$

For the ellipse, $\frac{r^2\cos^2\theta}{a^2}+\frac{r^2\sin^2\theta}{b^2}\le 1$ so its radial boundary is $r_E^2(\theta)=\frac{1}{\dfrac{\cos^2\theta}{a^2}+\dfrac{\sin^2\theta}{b^2}}.$

Now compare with the circles:

outer circle radius $a$
inner circle radius $b$

Since $\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}\ge \frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{a^2}=\frac{1}{a^2},$ we get $r_E^2(\theta)\le a^2 \implies r_E(\theta)\le a.$ So the ellipse lies entirely inside the circle $x^2+y^2\le a^2$ .

Also, $\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}\le \frac{\cos^2\theta}{b^2}+\frac{\sin^2\theta}{b^2}=\frac{1}{b^2},$ so $r_E^2(\theta)\ge b^2 \implies r_E(\theta)\ge b.$ Thus the circle $x^2+y^2\le b^2$ lies entirely inside the ellipse.

Therefore:

$C_a \cap E^c$ is simply the part of the big circle outside the ellipse, so its area is $\pi a^2-\pi ab.$
$C_b \cap E$ is simply the whole smaller circle, so its area is $\pi b^2.$

Use the given areas

From the second condition: $\pi b^2=18\pi$ so $b^2=18.$

From the first condition: $\pi a^2-\pi ab=30\pi$ so $a^2-ab=30.$

Also, area of ellipse is $\pi ab$ .

Find $a$ using $b=\sqrt{18}=3\sqrt2$

We have $a^2-ab=30.$ Substitute $b=3\sqrt2$ : $a^2-3\sqrt2\,a-30=0.$

Solve:

=\frac{3\sqrt2\pm\sqrt{138}}{2}.$$ Since $a>0$, take the positive root. But we actually only need $(a-b)^2$. --- 5. **Compute $(a-b)^2$ directly** Let $$d=a-b.$$ Then $a=b+d$. Use $$a^2-ab=30.$$ Substitute $a=b+d$: $$ (b+d)^2-b(b+d)=30$$ $$ b^2+2bd+d^2-b^2-bd=30$$ $$ bd+d^2=30$$ $$ d(b+d)=30.$$ But $b+d=a$, so this is again $ad=30$. Now use $b^2=18$ and test whether the given answer 12 is consistent: Compute $$(a-b)^2=d^2=a^2+b^2-2ab.$$ From $a^2-ab=30$, we have $$a^2=ab+30.$$ So $$d^2=(ab+30)+18-2ab=48-ab.$$ Thus we need $ab$. Since $a$ satisfies $a^2-ab=30$ and $b^2=18$, solve for $ab$ more neatly: Let $t=ab$. Then $$a=\frac{t}{b}.$$ Plug into $a^2-ab=30$: $$\frac{t^2}{b^2}-t=30.$$ Using $b^2=18$, $$\frac{t^2}{18}-t=30$$ $$t^2-18t-540=0$$ $$t=9\pm 3\sqrt{69}.$$ Since $t=ab>0$, take $$ab=9+3\sqrt{69}.$$ Then $$(a-b)^2=48-(9+3\sqrt{69})=39-3\sqrt{69},$$ which is not an integer. This suggests the intended expression in the question is likely $(a-b^2)$ or more probably $(a-b)^2$, and we should carefully re-check the geometry. --- 6. **Better geometric interpretation via set differences** Because ellipse is inside $C_a$, indeed $$\text{Area}(C_a\cap E^c)=\text{Area}(C_a)-\text{Area}(E)=\pi a^2-\pi ab=30\pi.$$ Because $C_b$ is inside the ellipse, $$\text{Area}(C_b\cap E)=\text{Area}(C_b)=\pi b^2=18\pi,$$ so $b^2=18$ is correct. Then $$a^2-ab=30.$$ This does not lead to an integer value for $(a-b)^2$. Hence the only natural intended quantity is likely $$ (a-b^2) $$ or perhaps $a-b$, but neither matches the stored key directly from the printed statement. Let us instead inspect whether the expression means $$ (a-b)2 $$ as in $2(a-b)$, a common OCR/formatting issue. From $$a^2-ab=30 \implies a(a-b)=30.$$ If $2(a-b)=12$, then $a-b=6$, giving $$a=\frac{30}{6}=5,$$ and then $$b=a-6=-1,$$ invalid. So this is impossible. Now try the stored answer 12 for $(a-b)^2$: $$a-b=2\sqrt3.$$ Then with $b^2=18$ we get $b=3\sqrt2$, so $$a=3\sqrt2+2\sqrt3,$$ and then $$a^2-ab \ne 30,$$ so this also fails. Therefore the stored answer does **not** agree with the mathematics of the given statement. --- 7. **Actual value from the given data** From $$b^2=18, \qquad a^2-ab=30,$$ we solve for $a-b$. Let $d=a-b$. Then $$a=b+d.$$ So $$a^2-ab=(b+d)^2-b(b+d)=bd+d^2=30.$$ With $b=3\sqrt2$, $$d^2+3\sqrt2\,d-30=0.$$ Hence $$d=\frac{-3\sqrt2+\sqrt{138}}{2}>0.$$ Therefore $$(a-b)^2=d^2= \left(\frac{\sqrt{138}-3\sqrt2}{2}\right)^2 =39-3\sqrt{69}.$$ So the value implied by the question is $$\boxed{39-3\sqrt{69}}.$$ This is not an integer, so the question/key appears inconsistent.

More from Area Under the Curves

View written solutionFree

Correct answer: 39-3\SQRT{69}

Interpret the two regions

We have:

Circle $C_a: x^2+y^2\le a^2$
Circle $C_b: x^2+y^2\le b^2$
Ellipse $E: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\le 1$

with $a>b>0$ .

We are given: $\text{Area}\big(C_a \cap E^c\big)=30\pi$ and $\text{Area}\big(C_b \cap E\big)=18\pi.$

We need $(a-b)^2$ .

Compare the circle and ellipse

Write both in polar form: $x=r\cos\theta,\qquad y=r\sin\theta.$

For the ellipse, $\frac{r^2\cos^2\theta}{a^2}+\frac{r^2\sin^2\theta}{b^2}\le 1$ so its radial boundary is $r_E^2(\theta)=\frac{1}{\dfrac{\cos^2\theta}{a^2}+\dfrac{\sin^2\theta}{b^2}}.$

Now compare with the circles:

outer circle radius $a$
inner circle radius $b$

Therefore:

$C_a \cap E^c$ is simply the part of the big circle outside the ellipse, so its area is $\pi a^2-\pi ab.$
$C_b \cap E$ is simply the whole smaller circle, so its area is $\pi b^2.$

Use the given areas

From the second condition: $\pi b^2=18\pi$ so $b^2=18.$

From the first condition: $\pi a^2-\pi ab=30\pi$ so $a^2-ab=30.$

Also, area of ellipse is $\pi ab$ .

Find $a$ using $b=\sqrt{18}=3\sqrt2$

We have $a^2-ab=30.$ Substitute $b=3\sqrt2$ : $a^2-3\sqrt2\,a-30=0.$

Solve:

=\frac{3\sqrt2\pm\sqrt{138}}{2}.$$ Since $a>0$, take the positive root. But we actually only need $(a-b)^2$. --- 5. **Compute $(a-b)^2$ directly** Let $$d=a-b.$$ Then $a=b+d$. Use $$a^2-ab=30.$$ Substitute $a=b+d$: $$ (b+d)^2-b(b+d)=30$$ $$ b^2+2bd+d^2-b^2-bd=30$$ $$ bd+d^2=30$$ $$ d(b+d)=30.$$ But $b+d=a$, so this is again $ad=30$. Now use $b^2=18$ and test whether the given answer 12 is consistent: Compute $$(a-b)^2=d^2=a^2+b^2-2ab.$$ From $a^2-ab=30$, we have $$a^2=ab+30.$$ So $$d^2=(ab+30)+18-2ab=48-ab.$$ Thus we need $ab$. Since $a$ satisfies $a^2-ab=30$ and $b^2=18$, solve for $ab$ more neatly: Let $t=ab$. Then $$a=\frac{t}{b}.$$ Plug into $a^2-ab=30$: $$\frac{t^2}{b^2}-t=30.$$ Using $b^2=18$, $$\frac{t^2}{18}-t=30$$ $$t^2-18t-540=0$$ $$t=9\pm 3\sqrt{69}.$$ Since $t=ab>0$, take $$ab=9+3\sqrt{69}.$$ Then $$(a-b)^2=48-(9+3\sqrt{69})=39-3\sqrt{69},$$ which is not an integer. This suggests the intended expression in the question is likely $(a-b^2)$ or more probably $(a-b)^2$, and we should carefully re-check the geometry. --- 6. **Better geometric interpretation via set differences** Because ellipse is inside $C_a$, indeed $$\text{Area}(C_a\cap E^c)=\text{Area}(C_a)-\text{Area}(E)=\pi a^2-\pi ab=30\pi.$$ Because $C_b$ is inside the ellipse, $$\text{Area}(C_b\cap E)=\text{Area}(C_b)=\pi b^2=18\pi,$$ so $b^2=18$ is correct. Then $$a^2-ab=30.$$ This does not lead to an integer value for $(a-b)^2$. Hence the only natural intended quantity is likely $$ (a-b^2) $$ or perhaps $a-b$, but neither matches the stored key directly from the printed statement. Let us instead inspect whether the expression means $$ (a-b)2 $$ as in $2(a-b)$, a common OCR/formatting issue. From $$a^2-ab=30 \implies a(a-b)=30.$$ If $2(a-b)=12$, then $a-b=6$, giving $$a=\frac{30}{6}=5,$$ and then $$b=a-6=-1,$$ invalid. So this is impossible. Now try the stored answer 12 for $(a-b)^2$: $$a-b=2\sqrt3.$$ Then with $b^2=18$ we get $b=3\sqrt2$, so $$a=3\sqrt2+2\sqrt3,$$ and then $$a^2-ab \ne 30,$$ so this also fails. Therefore the stored answer does **not** agree with the mathematics of the given statement. --- 7. **Actual value from the given data** From $$b^2=18, \qquad a^2-ab=30,$$ we solve for $a-b$. Let $d=a-b$. Then $$a=b+d.$$ So $$a^2-ab=(b+d)^2-b(b+d)=bd+d^2=30.$$ With $b=3\sqrt2$, $$d^2+3\sqrt2\,d-30=0.$$ Hence $$d=\frac{-3\sqrt2+\sqrt{138}}{2}>0.$$ Therefore $$(a-b)^2=d^2= \left(\frac{\sqrt{138}-3\sqrt2}{2}\right)^2 =39-3\sqrt{69}.$$ So the value implied by the question is $$\boxed{39-3\sqrt{69}}.$$ This is not an integer, so the question/key appears inconsistent.

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