f(x)=x2+9g(x)=x−9xa=f(g(10))=f(10−910)=f(10)=109b=g(f(3))=g(9+9)=g(18)=918=2E:109x2+2y2=1 e2=1−1092=109107ℓ=1092(2)=10948e2+ℓ2=1098(107)+10916=8
JEE Main 2024 — Mathematics Coordinate Geometry
Let f(x)=x2+9,g(x)=x−9x and a=f∘g(10),b=g∘f(3). If e and l denote the eccentricity and the length of the latus rectum of the ellipse ax2+by2=1, then 8e2+l2 is equal to.
Held on 9 Apr 2024 · Verified 6 Jul 2026.
8
16
6
12
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.