Let nCr−1=28,nCr=56{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}=28,{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=56nCr−1=28,nCr=56 and nCr+1=70{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}=70nCr+1=70. Let A(4cost,4sint),B(2sint,−2cost)\mathrm{A}(4 \cos t, 4 \sin t), \mathrm{B}(2 \sin t,-2 \cos \mathrm{t})A(4cost,4sint),B(2sint,−2cost) and C(3r−n,r2−n−1)C\left(3 r-n, r^2-n-1\right)C(3r−n,r2−n−1) be the vertices of a triangle ABCA B CABC, where ttt is a parameter. If (3x−1)2+(3y)2(3 x-1)^2+(3 y)^2(3x−1)2+(3y)2 =α=\alpha=α, is the locus of the centroid of triangle ABC , then α\alphaα equals
Held on 28 Jan 2025 · Verified 6 Jul 2026.
6
18
8
20
Sign in to track your attempts and accuracy.
nCr−1=28nCr=56nCr+1=70nCr−1nCr=2856⇒rn−r+1=12nCrnCr+1=5670⇒r+1n−r=7056}n=8=3\left.\begin{array}{l}{ }^n C_{r-1}=28 \\ { }^n C_r=56 \\ { }^n C_{r+1}=70 \\ \begin{array}{l}{ }^n C_{r-1} \\ { }^n C_r\end{array}=\frac{28}{56} \Rightarrow \frac{r}{n-r+1}=\frac{1}{2} \\ \frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{56}{70} \Rightarrow \frac{r+1}{n-r}=\frac{70}{56}\end{array}\right\} n=8=3nCr−1=28nCr=56nCr+1=70nCr−1nCr=5628⇒n−r+1r=21nCr+1nCr=7056⇒n−rr+1=5670⎭⎬⎫n=8=3
h=4cost+2sint+13k=4sint−2cost3h=\frac{4 \cos t+2 \sin t+1}{3} \quad k=\frac{4 \sin t-2 \cos t}{3}h=34cost+2sint+1k=34sint−2cost 3h−1=4cost+2sint3 h-1=4 \cos t+2 \sin t3h−1=4cost+2sint 3k−1=4sint−2cost3 k-1=4 \sin t-2 \cos t3k−1=4sint−2cost $\begin{aligned}& (1)^2+(2)^2 \& (3 h-1)^2+(3 k)^2=20\end{aligned}Locusofcentroid:Locus of centroid:Locusofcentroid:(3 x-1)^2+(3 y)^2=20\Rightarrow \alpha=20$
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
The distance between the points (3, 4) and (6, 8) is:
If the chord joining the points $\mathrm{P}_{1}\left(x_{1}, y_{1}\right)$ and $\mathrm{P}_{2}\left(x_{2}, y_{2}\right)$ on the parabola $y^{2}=12 x$ subtends a right angle at the vertex of the parabola, then $x_{1} x_{2}-y_{1} y_{2}$ is equal to
The distance between the parallel lines 3x + 4y - 7 = 0 and 3x + 4y + 8 = 0 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 :
1966 questions
1411 questions
485 questions
389 questions
224 questions
Work through every JEE Main Coordinate Geometry PYQ, year by year.