A pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve x2 + y2 = 4 with x + y = a. The set containing the value of 'a' is

Question

A pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve x2 + y2 = 4 with x + y = a. The set containing the value of 'a' is (A)  -2, 2  (B)  -3, 3  (C)  -4, 4  (D)  -5, 5

Tardigrade Admin · Accepted Answer

To make the given curves x2+y2=4 and x+y=a homogenous. ∴ x2+y2-4 (x +y2/x)=0 ⇒ a2(x2+y2)-4(x2+y2+2 x y)=0 ⇒ x2(a2-4)+y2(a2-4)-8 x y=0 Since, this is a perpendicular pair of straight lines. ∴ a2-4+a2-4=0 ⇒ a2=4 ⇒ a=± 2 Hence, required set of a is -2,2 .

Q. A pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve $x^{2} + y^{2} = 4$ with $x + y = a$ . The set containing the value of ' $a$ ' is

${- 2, 2}$

${- 3, 3}$

${- 4, 4}$

${- 5, 5}$

Q. A pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve x2+y2=4 with x+y=a. The set containing the value of 'a' is

{−2,2}

{−3,3}

{−4,4}

{−5,5}

To make the given curves x2+y2=4 and x+y=a homogenous. ∴x2+y2−4xx+y2​=0 ⇒a2(x2+y2)−4(x2+y2+2xy)=0 ⇒x2(a2−4)+y2(a2−4)−8xy=0 Since, this is a perpendicular pair of straight lines. ∴a2−4+a2−4=0 ⇒a2=4 ⇒a=±2 Hence, required set of a is {−2,2}.

Q. A pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve $x^{2} + y^{2} = 4$ with $x + y = a$ . The set containing the value of ' $a$ ' is

${- 2, 2}$

${- 3, 3}$

${- 4, 4}$

${- 5, 5}$