Question

The equation \(\frac{{{x^2}}}{{2 - r}} + \frac{{{y^2}}}{{r - 6}} + 1 = 0\) represents an ellipse if
1. r > 2
2. r > 6
3. 2 < r < 6
4. None of these

Answer 1

Correct Answer - Option 3 : 2 < r < 6

Concept:

Ellipse:  It is a locus of a point moving in such a way that the sums of its distances from two fixed points are a constant. The fixed points are known as the foci. 

The standard form of equation of a conic section is

Ax² + Bxy + Cy² + Dx + Ey + F = 0

Where A, B, C, D, E, F are real numbers and A ≠ 0, B ≠ 0, C ≠ 0.

•  B² – 4AC < 0, then the conic section is an ellipse.
•  B² – 4AC = 0, then the conic section is a parabola.
•  B² – 4AC > 0, then the conic section is a hyperbola.

Calculation:

Given that,

 \(\dfrac{x^2}{2-r} + \dfrac{y^2}{r-6}+1=0\)

Equating the equation of the ellipse with the second-degree equation

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, we get

 \(A=\dfrac{1}{2-r}, B = 0, C = \dfrac{1}{r-6}, D = 0, E = 0\ \ and\ \ F\ =\ 0\) We know that, for the second-degree equation to represent an ellipse, the coefficients must satisfy the discriminant condition

B² - 4AC < 0 and also A ≠ C

\(⇒ (0)^2 - 4 \left(\dfrac{1}{2-r}\right) \left(\dfrac{1}{r-6}\right) < 0\)

\(⇒ -4 \left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-6}\right)<0\)

\(⇒ \left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-6}\right)>0\)

⇒ (2 - r) (r - 6) < 0

⇒ (r - 2) (r - 6) > 0

⇒ r > 2 and r < 6

⇒ 2 < r < 6

Hence, above equation will represents ellipse if 2 < r < 6.