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  • Let a line y=m x(m>0) intersect the parabola, at a point P, other than the origin. Let

Let a line y=m x(m>0) intersect the parabola, \mathrm{y^2=x} at a point P, other than the origin. Let the tangent to it at P meet the x-axis at the point Q. If area \mathrm{(\triangle O P Q)=4} sq. units, then m is equal to

Option: 1

0.5


Option: 2

0.6


Option: 3

0.7


Option: 4

0.8


Answers (1)

best_answer

The given curve is \mathrm{y2 = x}               ....[i]

\mathrm{\text { Since } P \text { is on curve (i) }}

\mathrm{\therefore \quad \text { Coordinates of } P \text { are }\left(t^2, t\right)}

\mathrm{\text { Equation of tangent at } P\left(t^2, t\right) \text { is }}

\mathrm{2 t y=x+t^2}                 ....[ii]

Since equation (ii) meet at
x-axis, therefore y = 0

\mathrm{\text { So, } Q\left(-t^2, 0\right) \text {. }}

\mathrm{\text { Since, } \operatorname{area}(\triangle O P Q)=4 \text { sq. units [Given] }}

\mathrm{\begin{aligned} & \therefore \quad \frac{1}{2}\left\|\begin{array}{ccc} 0 & 0 & 1 \\ t^2 & t & 1 \\ -t^2 & 0 & 1 \end{array}\right\|=4 \\ & \Rightarrow\left|t^3\right|=8 \Rightarrow t= \pm 2 \end{aligned}}

\mathrm{\therefore \quad \text { Equation of } O P \text { is } y=\frac{2}{4}(x) \Rightarrow m=\frac{1}{2}=0.5}

 

Posted by

Devendra Khairwa

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