The sum of squares of all possible values of k , for which area of the region bounded by the parabolas 2 y^2=k x and ky^2=2(y-x) is maximum, is equal to :

Area Under the Curves PYQ — JEE Main 2024 · 1 Feb · Shift 2 · Q52 with Solution

View written solutionFree

Correct answer: 8

Given parabolas

We are given $2y^2 = kx \quad \text{and} \quad ky^2 = 2(y-x).$

Rewrite both in the form $x=f(y)$ .

From the first parabola: $x = \frac{2y^2}{k}.$

From the second parabola:

\implies 2x = 2y - ky^2 \implies x = y - \frac{k}{2}y^2.$$ So the curves are $$x_1 = \frac{2y^2}{k}, \qquad x_2 = y - \frac{k}{2}y^2.$$ --- 2. **Points of intersection** At intersection, $$\frac{2y^2}{k} = y - \frac{k}{2}y^2.$$ Multiply by $2k$: $$4y^2 = 2ky - k^2y^2.$$ Rearrange: $$(k^2+4)y^2 - 2ky = 0.$$ Factor: $$y\big((k^2+4)y - 2k\big)=0.$$ Hence, $$y=0 \quad \text{or} \quad y=\frac{2k}{k^2+4}.$$ So the enclosed area exists when these two $y$-values are distinct, i.e. $k\neq 0$. --- 3. **Area between the curves** The area is $$A = \int (x_{\text{right}}-x_{\text{left}})\,dy.$$ Now $$x_2-x_1 = y - \frac{k}{2}y^2 - \frac{2y^2}{k} = y - \left(\frac{k}{2}+\frac{2}{k}\right)y^2.$$ For $k>0$, the second intersection is positive, and area is $$A=\int_0^{\frac{2k}{k^2+4}} \left[y - \left(\frac{k}{2}+\frac{2}{k}\right)y^2\right]dy.$$ Let $$\alpha=\frac{k}{2}+\frac{2}{k}=\frac{k^2+4}{2k}.$$ Then $$A=\int_0^{\frac{2k}{k^2+4}} (y-\alpha y^2)\,dy = \left[\frac{y^2}{2}-\frac{\alpha y^3}{3}\right]_0^{\frac{2k}{k^2+4}}.$$ Let $$b=\frac{2k}{k^2+4}.$$ Since $$\alpha b = \frac{k^2+4}{2k}\cdot \frac{2k}{k^2+4}=1,$$ we get $$A=\frac{b^2}{2}-\frac{b^2}{3}=\frac{b^2}{6}.$$ Thus $$A=\frac{1}{6}\left(\frac{2k}{k^2+4}\right)^2 =\frac{2k^2}{3(k^2+4)^2}.$$ For $k<0$, the integration limits reverse, but area remains the positive magnitude, so the same formula for area holds: $$A(k)=\frac{2k^2}{3(k^2+4)^2}.$$ --- 4. **Maximize the area** We need to maximize $$A(k)=\frac{2k^2}{3(k^2+4)^2}.$$ Since $\frac{2}{3}$ is constant, maximize $$f(k)=\frac{k^2}{(k^2+4)^2}.$$ Put $$t=k^2 \quad (t\ge 0).$$ Then maximize $$g(t)=\frac{t}{(t+4)^2}.$$ Differentiate: $$g'(t)=\frac{(t+4)^2-2t(t+4)}{(t+4)^4} =\frac{(t+4)(4-t)}{(t+4)^4} =\frac{4-t}{(t+4)^3}.$$ Set $g'(t)=0$: $$4-t=0 \implies t=4.$$ Hence, $$k^2=4 \implies k=\pm 2.$$ --- 5. **Sum of squares of all possible values of $k$** Possible values are $$k=2,\,-2.$$ Their squares are $$2^2=4, \qquad (-2)^2=4.$$ So the required sum is $$4+4=8.$$ --- 6. **Comparison with stored answer** Derived answer = $8$. Stored correct answer = $8$. They agree.

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View written solutionFree

Correct answer: 8

Given parabolas

We are given $2y^2 = kx \quad \text{and} \quad ky^2 = 2(y-x).$

Rewrite both in the form $x=f(y)$ .

From the first parabola: $x = \frac{2y^2}{k}.$

From the second parabola:

\implies 2x = 2y - ky^2 \implies x = y - \frac{k}{2}y^2.$$ So the curves are $$x_1 = \frac{2y^2}{k}, \qquad x_2 = y - \frac{k}{2}y^2.$$ --- 2. **Points of intersection** At intersection, $$\frac{2y^2}{k} = y - \frac{k}{2}y^2.$$ Multiply by $2k$: $$4y^2 = 2ky - k^2y^2.$$ Rearrange: $$(k^2+4)y^2 - 2ky = 0.$$ Factor: $$y\big((k^2+4)y - 2k\big)=0.$$ Hence, $$y=0 \quad \text{or} \quad y=\frac{2k}{k^2+4}.$$ So the enclosed area exists when these two $y$-values are distinct, i.e. $k\neq 0$. --- 3. **Area between the curves** The area is $$A = \int (x_{\text{right}}-x_{\text{left}})\,dy.$$ Now $$x_2-x_1 = y - \frac{k}{2}y^2 - \frac{2y^2}{k} = y - \left(\frac{k}{2}+\frac{2}{k}\right)y^2.$$ For $k>0$, the second intersection is positive, and area is $$A=\int_0^{\frac{2k}{k^2+4}} \left[y - \left(\frac{k}{2}+\frac{2}{k}\right)y^2\right]dy.$$ Let $$\alpha=\frac{k}{2}+\frac{2}{k}=\frac{k^2+4}{2k}.$$ Then $$A=\int_0^{\frac{2k}{k^2+4}} (y-\alpha y^2)\,dy = \left[\frac{y^2}{2}-\frac{\alpha y^3}{3}\right]_0^{\frac{2k}{k^2+4}}.$$ Let $$b=\frac{2k}{k^2+4}.$$ Since $$\alpha b = \frac{k^2+4}{2k}\cdot \frac{2k}{k^2+4}=1,$$ we get $$A=\frac{b^2}{2}-\frac{b^2}{3}=\frac{b^2}{6}.$$ Thus $$A=\frac{1}{6}\left(\frac{2k}{k^2+4}\right)^2 =\frac{2k^2}{3(k^2+4)^2}.$$ For $k<0$, the integration limits reverse, but area remains the positive magnitude, so the same formula for area holds: $$A(k)=\frac{2k^2}{3(k^2+4)^2}.$$ --- 4. **Maximize the area** We need to maximize $$A(k)=\frac{2k^2}{3(k^2+4)^2}.$$ Since $\frac{2}{3}$ is constant, maximize $$f(k)=\frac{k^2}{(k^2+4)^2}.$$ Put $$t=k^2 \quad (t\ge 0).$$ Then maximize $$g(t)=\frac{t}{(t+4)^2}.$$ Differentiate: $$g'(t)=\frac{(t+4)^2-2t(t+4)}{(t+4)^4} =\frac{(t+4)(4-t)}{(t+4)^4} =\frac{4-t}{(t+4)^3}.$$ Set $g'(t)=0$: $$4-t=0 \implies t=4.$$ Hence, $$k^2=4 \implies k=\pm 2.$$ --- 5. **Sum of squares of all possible values of $k$** Possible values are $$k=2,\,-2.$$ Their squares are $$2^2=4, \qquad (-2)^2=4.$$ So the required sum is $$4+4=8.$$ --- 6. **Comparison with stored answer** Derived answer = $8$. Stored correct answer = $8$. They agree.

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