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Area of the triangle formed by the lines  \mathrm {y^2-9 x y+18 x^2=0} and \mathrm{y = a} is 

 

Option: 1

\frac{27}{4}


Option: 2

0


Option: 3

\frac{a}{3}


Option: 4

\frac{a^2}{12}


Answers (1)

best_answer

Given pair of lines is \mathrm{y^2-9 x y+18 x^2=0} _________(I)

Or \mathrm{(y-3 x)(y-6 x)=0}

Hence given lines are \mathrm{y-3x=0} ---------(II)

                                    \mathrm{y-6x=0} ---------(III)

                and  \mathrm{y=a} -----------(IV)

Vertices of triangle formed are  \mathrm{(0,0),\left(\frac{a}{3}, a\right),\left(\frac{a}{6}, a\right)}

Area of the triangle =  \mathrm{e}=\frac{1}{2}\left|\left(\frac{a}{3} \cdot a-a \cdot \frac{a}{6}\right)\right|=\frac{a^2}{12}

 

 

Posted by

Gautam harsolia

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