JEE Math Practice Question
If the midpoint of a chord of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is $(\sqrt{2}, 4 / 3)$, and the length of the chord is $\frac{2 \sqrt{\alpha}}{3}$, then $\alpha$ is :
- A.26
- B.18
- C.22
- D.20
Want to know if you got it right?
Answer this and thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.
Practice JEE free →More JEE Math questions
- What is the total number of distinct x \in \mathbb{R} for which \left|\begin{array}{ccc}x …
- Let m be the smallest positive integer such that the coefficient of x^{2} in the expansion…
- What is the total number of distinct x \in[0,1] for which \int_{0}^{x} \frac{t^{2}}{1+t^{4…
- Let \alpha, \beta \in \mathbb{R} be such that \lim _{x \rightarrow 0} \frac{x^{2} \sin (\b…
- Let z=\frac{-1+\sqrt{3} i}{2}, where i=\sqrt{-1}, and r, s \in\{1,2,3\}. Let P=\left[\begi…
- For how many values of p, the circle x^{2}+y^{2}+2 x+4 y-p=0 and the coordinate axes have …
- Let f: \mathbb{R} \rightarrow \mathbb{R} be a differentiable function such that f(0)=0, f\…
- For a real number \alpha, if the system \[ \left[\begin{array}{ccc} 1 & \alpha & \alpha^{2…