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TS EAMCET 4 May 2018 Shift 2 Solved Paper

Section: Mathematics

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Question : 56 of 160

Marks: +1, -0

If the straight line

8x+3\sqrt{2}y=36

touches the ellipse

\frac{x^{2}}{9}+\frac{y^{2}}{4}=2

(a, b)

a+\sqrt{2}b=

$\frac{36}{5\sqrt{2}}$
$\frac{8}{3}$
$\frac{12+2\sqrt{2}}{3}$
$\frac{16}{3}$

Solution:

Consider the given equation,

8x+3\sqrt{2}y=36

And,

\frac{x^{2}}{9}+\frac{y^{2}}{4}=2

\frac{x x_{1}}{9}+\frac{y y_{1}}{4}-2=0

........(I)

And,

8x+3\sqrt{2}y-36=0

......(II)

Since equation (I) and (II) represents same line so,

\frac{x_{1}}{9(8)}=\frac{y_{1}}{4(3\sqrt{2})}=\frac{1}{18}

From above,

x_{1}=4=a

And,

y_{1}=\frac{12\sqrt{2}}{18}=b

This implies,

a+\sqrt{2}b=4+\sqrt{2}\frac{2\sqrt{2}}{3}

\frac{12+4}{3}=\frac{16}{3}

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