JEE Mains · Maths · STD 11 - 9. straight line
Let \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}=28,{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=56\) and \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}=70\). Let \(\mathrm{A}(4 \cos t, 4 \sin t), \mathrm{B}(2 \sin t,-2 \cos \mathrm{t})\) and \(C\left(3 r-n, r^2-n-1\right)\) be the vertices of a triangle \(A B C\), where \(t\) is a parameter. If \((3 x-1)^2+(3 y)^2\) \(=\alpha\), is the locus of the centroid of triangle ABC , then \(\alpha\) equals
- A 6
- B 18
- C 8
- D 20
Answer & Solution
Correct Answer
(D) 20
Step-by-step Solution
Detailed explanation
\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…
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