JEE MainMathematicsEasyEllipse

If
$\frac{x^{2}}{r^{2} - r - 6} + \frac{y^{2}}{r^{2} - 6r + 5} = 1$ will
represents the ellipse, then set of all positive values of r is
ä djrh gS] rc r

Show answer and solution

\frac{x^{2}}{r^{2} - r - 6} + \frac{y^{2}}{r^{2} - 6r + 5} = 1

represents an ellipse then,

r^{2} - r - 6 > 0\ and\ r^{2} - 6r + 5 > 0;\ \ \ \ (r - 3)(r + 2) > 0

and

\ \ \ \ \ \ \ \ \ \ \ \ \ \ (r - 1)(r - 5) > 0r \in ( - \infty, - 2) \cup (3,\infty)\ and\ r \in ( - \infty, - 1) \cup (5,\infty)

Hence,\ r\ lies\ in\ the\ interval\ \Rightarrow r \in ( - \infty, - 2) \cup (5,\infty)

\frac{x^{2}}{r^{2} - r - 6} + \frac{y^{2}}{r^{2} - 6r + 5} = 1

nh?kZo`Ùk
gS rc

r^{2} - r - 6 > 0\

vkSj

r^{2} - 6r + 5 > 0;(r - 3)(r + 2) > 0

vr r vUrjky esa fLFkr

\Rightarrow r \in ( - \infty, - 2) \cup (5,\infty)

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx

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