The curve y=(|x|-1) operatornamesgn(x-1) divides (9 x2/64)+(4 y2/25)=(1/π) into two parts of areas A1 and A2 ( A 1

Question

The curve y=(|x|-1) operatornamesgn(x-1) divides (9 x2/64)+(4 y2/25)=(1/π) into two parts of areas A1 and A2 ( A 1< A 2), then [Note: operatornamesgn( k ) denotes signum function of k.] (A) A 1=(7/6) (B) A1: A2=7: 13 (C) A2=(13/6) (D) A 2- A 1=1

Tardigrade Admin · Accepted Answer

Area of an ellipse =π ab =(20/3), Area of triangle P Q R=1 ∴ A 1=(7/3) ; A 2=(13/3) [By symmetry area M = area N ]

Q. The curve $y = (∣ x ∣ - 1) sgn (x - 1)$ divides $\frac{9 x ^{2}}{64} + \frac{4 y ^{2}}{25} = \frac{1}{π}$ into two parts of areas $A_{1}$ and $A_{2}$ $(A_{1} < A_{2})$ , then
[Note: $sgn (k)$ denotes signum function of $k$ .]

$A_{1} = \frac{7}{6}$

$A_{1} : A_{2} = 7 : 13$

$A_{2} = \frac{13}{6}$

$A_{2} - A_{1} = 1$

Area of an ellipse $= πab = \frac{20}{3}$ , Area of triangle $PQR = 1$
$∴ A_{1} = \frac{7}{3}; A_{2} = \frac{13}{3}$ [By symmetry area $M =$ area $N$ ]

Q. The curve y=(∣x∣−1)sgn(x−1) divides 649x2​+254y2​=π1​ into two parts of areas A1​ and A2​ (A1​<A2​), then [Note: sgn(k) denotes signum function of k.]

A1​=67​

A1​:A2​=7:13

A2​=613​

A2​−A1​=1