
Homogenising
x2+2y2−2(x+y)2=0
⇒−x2−4xy=0⇒x2+4xy=0
Lines are x=0 and y=−4x
∴ Angle between lines =2π+tan−141
option (3)
JEE Main 2021 — Mathematics Coordinate Geometry
If the curve x2+2y2=2 intersects the line x+y=1 at two points P and Q, then the angle subtended by the line segment PQ at the origin is
Held on 25 Feb 2021 · Verified 6 Jul 2026.
2π−tan−1(31)
2π+tan−1(31)
2π+tan−1(41)
2π−tan−1(41)
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Suppose that two chords, drawn from the point $(1, 2)$ on the circle $x^2 + y^2 + x - 3y = 0$ are bisected by the $y$-axis. If the other ends of these chords are $R$ and $S$, and the mid point of the line segment $RS$ is $(\alpha, \beta)$, then $6(\alpha + \beta)$ is equal to:
Let $\dfrac{x^2}{f(a^2+7a+3)} + \dfrac{y^2}{f(3a+15)} = 1$ represent an ellipse with major axis along $y$-axis, where $f$ is a strictly decreasing positive function on $\mathbb{R}$. If the set of all possible values of $a$ is $\mathbb{R} - [\alpha, \beta]$, then $\alpha^2+\beta^2$ is equal to:
Let the vertex $A$ of a triangle $ABC$ be $(1, 2)$, and the mid-point of the side $AB$ be $(5, -1)$. If the centroid of this triangle is $(3, 4)$ and its circumcenter is $(\alpha, \beta)$, then $21(\alpha + \beta)$ is equal to:
Let O be the origin, and P and Q be two points on the rectangular hyperbola $xy = 12$ such that the mid point of the line segment PQ is $\left(\dfrac{1}{2}, -\dfrac{1}{2}\right)$. Then the area of the triangle OPQ equals:
If the line $\alpha x+4 y=\sqrt{7}$, where $\alpha \in \mathbf{R}$, touches the ellipse $3 x^{2}+4 y^{2}=1$ at the point P in the first quadrant, then one of the focal distances of $P$ is :
Work through every JEE Main Coordinate Geometry PYQ, year by year.