A pair of perpendicular straight lines passes | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To make the given curves $x^2+y^2=4$ and $x+y=a$ homogeneous,we write the equation of the pair of lines passing through the origin and the interse

Vedclass · Accepted Answer

$\{-2, 2\}$

Vedclass · Answer

(A) To make the given curves $x^2+y^2=4$ and $x+y=a$ homogeneous,we write the equation of the pair of lines passing through the origin and the intersection points as follows:$x^2+y^2-4\left(\frac{x+y}{a}\right)^2=0$$a^2(x^2+y^2)-4(x^2+y^2+2xy)=0$$x^2(a^2-4) - 8xy + y^2(a^2-4)=0$Since this represents a pair of perpendicular straight lines,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$(a^2-4) + (a^2-4) = 0$$2(a^2-4) = 0$$a^2 = 4$$a = \pm 2$Hence,the required set of $a$ is $\{-2, 2\}$.

$A$ 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

Similar Questions

In eubacteria,a cellular component that resembles eukaryotic cell is

The radii of curvature of the faces of a double convex lens are $10\, cm$ and $15\, cm$. Its focal length is $12\, cm$. What is the refractive index of the glass?

What is the process of breeding crops to increase their levels of minerals,vitamins,and proteins called?

Water in a river $20 \ m$ deep is flowing at a speed of $10 \ ms^{-1}$. The shearing stress between the horizontal layers of water in the river in $Nm^{-2}$ is (Coefficient of viscosity of water $= 10^{-3} \ SI$ units).

$A$ magnetic needle lying parallel to a magnetic field is turned through $60^{\circ}$. The work done on it is $W$. The torque required to maintain the magnetic needle in the position mentioned above is

Vedclass Test Series

Exam Paper Generator

Online Exam Module