If the area enclosed between the curves y = kx2 and x=ky2, (k 0), is 1 square unit. Then k is:

Question

If the area enclosed between the curves y = kx2 and x=ky2, (k > 0), is 1 square unit. Then k is: (A) (1/√3) (B) (2/√3) (C) (√3/2) (D) √3

Tardigrade Admin · Accepted Answer

Area bounded by y2 = 4ax x2 = 4by, a, b ≠ 0 is | (16ab/3) | by using formula: 4a = (1/k) = 4b, k > 0 Area = |(16 . (1/4k) . (1/4k)/3)| = 1 ⇒ k2 = (1/3) ⇒ k = (1/√3)

Q. If the area enclosed between the curves $y = k x^{2}$ and $x = k y^{2}$ , $(k > 0)$ , is $1$ square unit. Then $k$ is:

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{3}{2}$

$3$

Area bounded by $y^{2} = 4 a x & x^{2} = 4 b y, a, b \neq = 0$ is $∣ ∣ \frac{16 ab}{3} ∣ ∣$
by using formula : $4 a = \frac{1}{k} = 4 b, k > 0$
Area $= ∣ ∣ \frac{16. \frac{1}{4 k} . \frac{1}{4 k}}{3} ∣ ∣ = 1$
$\Rightarrow k^{2} = \frac{1}{3}$
$\Rightarrow k = \frac{1}{3}$

Q. If the area enclosed between the curves y=kx2 and x=ky2, (k>0), is 1 square unit. Then k is:

3​1​

3​2​

23​​

3​

Area bounded by y2=4ax&x2=4by,a,b=0 is ∣∣​316ab​∣∣​ by using formula : 4a=k1​=4b,k>0 Area =∣∣​316.4k1​.4k1​​∣∣​=1 ⇒k2=31​ ⇒k=3​1​

Q. If the area enclosed between the curves $y = k x^{2}$ and $x = k y^{2}$ , $(k > 0)$ , is $1$ square unit. Then $k$ is:

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{3}{2}$

$3$

Area bounded by $y^{2} = 4 a x & x^{2} = 4 b y, a, b \neq = 0$ is $∣ ∣ \frac{16 ab}{3} ∣ ∣$
by using formula : $4 a = \frac{1}{k} = 4 b, k > 0$
Area $= ∣ ∣ \frac{16. \frac{1}{4 k} . \frac{1}{4 k}}{3} ∣ ∣ = 1$
$\Rightarrow k^{2} = \frac{1}{3}$
$\Rightarrow k = \frac{1}{3}$