First slide

Ellipse

Question

$The equation of the ellipse whose axes are coincident with the coordinates axes and which touches the straight lines$
$3 x - 2 y - 20 = 0 and x + 6 y - 20 = 0 is$

Moderate

A

$\frac{x^{2}}{40} + \frac{y^{2}}{10} = 1$
B

$\frac{x^{2}}{5} + \frac{y^{2}}{8} = 1$
C

$\frac{x^{2}}{10} + \frac{y^{2}}{40} = 1$
D

$\frac{x^{2}}{40} + \frac{y^{2}}{30} = 1$

Solution

$Let the equation of the ellipse be \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$We know that the general equation of the tangent to the ellipse is$
$y = m x \pm \sqrt{a^{2} m^{2} + b^{2}} - - - - (i)$
$\begin{array}{l} Since 3 x - 2 y - 20 = 0 or y = \frac{3}{2} x - 10 \\ is tangent to the ellipse, comparing with (i), \end{array}$
$\begin{array}{l} m = \frac{3}{2} and a^{2} m^{2} + b^{2} = 100 \\ or a^{2} \times \frac{9}{4} + b^{2} = 100 \\ or 9 a^{2} + 4 b^{2} = 400 - - - - - (i i) \end{array}$
$Similarly, since x + 6 y - 20 = 0, i.e.,$
$y = - \frac{1}{6} x + \frac{10}{3}$
$is tangent to the ellipse, comparing with (i),$
$\begin{array}{l} m = \frac{1}{6} and a^{2} m^{2} + b^{2} = \frac{100}{9} \\ or \frac{a^{2}}{36} + b^{2} = \frac{100}{9} \\ or a^{2} + 36 b^{2} = 400 - - - - - (i i i) \end{array}$
$Solving (ii) and (iii), we get a^{2} = 40 and b^{2} = 10 .$
$Therefore, the required equation of the ellipse is$
$\frac{x^{2}}{40} + \frac{y^{2}}{10} = 1$

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