First slide

Ellipse

Question

$If the equation of the ellipse whose axes are coincident with the coordinate axes and$
$which touches the straight lines 3 x - 2 y - 20 = 0 and x + 6 y - 20 = 0 is \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, then a+b=$

Moderate

A

50
B

$5 \sqrt{10}$
C

$3 \sqrt{10}$
D

30

Solution

$Let the equation of the ellipse be \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \dots \dots \dots \dots \dots . (1)$
Equation of the tangent to the ellipse in slope form is
$y = m x \pm \sqrt{a^{2} m^{2} + b^{2}} \dots \dots \dots \dots \dots (2)$
$Given equation of the tangent is 3 x - 2 y - 20 = 0$
Compare (2) and (3)
$\begin{array}{l} \Rightarrow m = \frac{3}{2} and a^{2} m^{2} + b^{2} = 100 \\ \Rightarrow a^{2} (\frac{9}{4}) + b^{2} = 100 \\ \Rightarrow 9 a^{2} + 4 b^{2} = 400 \dots \dots \dots \dots \dots (4) \\ Given equation of the another tangent is x + 6 y - 20 = 0 \\ \Rightarrow y = - \frac{1}{6} x + \frac{10}{3} \dots \dots \dots \dots \dots \dots (5) \end{array}$
Compare (2) and (5)
$\begin{array}{l} We get m = - \frac{1}{6} and a^{2} m^{2} + b^{2} = \frac{100}{9} \\ \Rightarrow \frac{a^{2}}{36} + b^{2} = \frac{100}{9} \\ \Rightarrow a^{2} + 36 b^{2} = 400 \dots \dots (6) \end{array}$
Now solving equations (4) & (6)
$we get a^{2} = 40 and b^{2} = 10$
$\begin{array}{l} \Rightarrow a = \sqrt{40} = 2 \sqrt{10}, b = \sqrt{10} \\ \Rightarrow a + b = 3 \sqrt{10} \end{array}$
Therefore, the correct answer is (3).

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