
Area of △PQR
$\begin{aligned}
& =\frac{1}{2}(2 a)(a \sin \theta+b) \
& \therefore \text { maximum area }=a(a+b) \
& \quad=4(4+2 \sqrt{3})=8(2+\sqrt{3})
\end{aligned}$
JEE Main 2025 — Mathematics Coordinate Geometry
Let C be the circle of minimum area enclosing the ellipse E:a2x2+b2y2=1 with eccentricity 21 and foci (±2,0). Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 29 is parallel to the major axis of E and contains the point of intersection of E with the negative y-axis. Then the maximum area of the triangle PQR is :
Held on 3 Apr 2025 · Verified 6 Jul 2026.
6(3+2)
8(3+2)
62+3
82+3
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Let the image of parabola $x^{2}=4 y$, in the line $x-y=1$ be $(y+a)^{2}=b(x-c)$, $a, b, c \in \mathrm{~N}$. Then $a+b+c$ is equal to
Let the domain of the function $f(x)=\log _{3} \log _{5} \log _{7}\left(9 x-x^{2}-13\right)$ be the interval $(\mathrm{m}, \mathrm{n})$. Let the hyperbola $\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1$ have eccentricity $\frac{\mathrm{n}}{3}$ and the length of the latus rectum $\frac{8 \mathrm{~m}}{3}$. Then $\mathrm{b}^{2}-\mathrm{a}^{2}$ is equal to :
The distance between the points (3, 4) and (6, 8) is:
If P is a point on the circle $x^{2}+y^{2}=4, \mathrm{Q}$ is a point on the straight line $5 x+y+2=0$ and $x-y+1=0$ is the perpendicular bisector of PQ, then 13 times the sum of abscissa of all such points P is $\_\_\_\_$.
Let a point $A$ lie between the parallel lines $L_{1}$ and $L_{2}$ such that its distances from $L_{1}$ and $L_{2}$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle $A B C$, where the points $B$ and C lie on the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$, respectively, is :
Work through every JEE Main Coordinate Geometry PYQ, year by year.