The area (in sq. units) enclosed between the curve x=(1 - t2/1 + t2),y=(2 t/1 + t2),∀ t∈ R and the line y=x+1 above the line is

Question

The area (in sq. units) enclosed between the curve x=(1 - t2/1 + t2),y=(2 t/1 + t2),∀ t∈ R and the line y=x+1 above the line is (A) (π /4) (B) (1/2) (C) (3 π /4)+(1/2) (D) (π /4)-(1/2)

Tardigrade Admin · Accepted Answer

Let, t=tan θ ⇒ x=cos 2θ ,y=sinâ¡2θ ⇒ x2+y2=1 <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-re0t1mrltou9lfvp.jpg" /> Area of the shaded region is the difference of (1/4) of the circle and area of the triangle according to the figure. =(π /4)-(1/2)× 1× 1=(π /4)-(1/2) sq. units

Q. The area (in sq. units) enclosed between the curve $x = \frac{1 - t ^{2}}{1 + t ^{2}}, y = \frac{2 t}{1 + t ^{2}}, \forall t \in R$ and the line $y = x + 1$ above the line is

$\frac{π}{4}$

$\frac{1}{2}$

$\frac{3 π}{4} + \frac{1}{2}$

$\frac{π}{4} - \frac{1}{2}$

Q. The area (in sq. units) enclosed between the curve x=1+t21−t2​,y=1+t22t​,∀t∈R and the line y=x+1 above the line is

4π​

21​

43π​+21​

4π​−21​

Let, t=tanθ ⇒x=cos2θ,y=sin⁡2θ ⇒x2+y2=1 Area of the shaded region is the difference of 41​ of the circle and area of the triangle according to the figure. =4π​−21​×1×1=4π​−21​ sq. units

Q. The area (in sq. units) enclosed between the curve $x = \frac{1 - t ^{2}}{1 + t ^{2}}, y = \frac{2 t}{1 + t ^{2}}, \forall t \in R$ and the line $y = x + 1$ above the line is

$\frac{π}{4}$

$\frac{1}{2}$

$\frac{3 π}{4} + \frac{1}{2}$

$\frac{π}{4} - \frac{1}{2}$