10-2. Parabola, Ellipse, Hyperbola
hard

If $x^{2}+9 y^{2}-4 x+3=0, x, y \in R$, then $x$ and $y$ respectively lie in the intervals:

A

$\left[-\frac{1}{3}, \frac{1}{3}\right]$ and $\left[-\frac{1}{3}, \frac{1}{3}\right]$

B

$\left[-\frac{1}{3}, \frac{1}{3}\right]$ and $[1,3]$

C

$[1,3]$ and $[1,3]$

D

$[1,3]$ and $\left[-\frac{1}{3}, \frac{1}{3}\right]$

(JEE MAIN-2021)

Solution

 

$x^{2}+9 y^{2}-4 x+3=0$

$\left(x^{2}-4 x\right)+\left(9 y^{2}\right)+3=0$

$\left(x^{2}-4 x+4\right)+\left(9 y^{2}\right)+3-4=0$

$(x-2)^{2}+(3 y)^{2}=1$

$\frac{(x-2)^{2}}{(1)^{2}}+\frac{y^{2}}{\left(\frac{1}{3}\right)^{2}}=1$ (equation of an ellipse).

As it is equation of an ellipse, $x \,\&\, y$ can vary inside the ellipse.

So, $x-2 \in[-1,1]$ and $y \in\left[-\frac{1}{3}, \frac{1}{3}\right]$

$x \in[1,3] y \in\left[-\frac{1}{3}, \frac{1}{3}\right]$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.