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  • From a point A common tangents are drawn to the circle 

From a point A common tangents are drawn to the circle \mathrm{x^2+y^2=a^2 / 2}  and the parabola  \mathrm{y^2=4 a x}. Find the area of the quadrilateral formed by the common tangents, the chords of contact of the point A, w.r.t. the circle and the parabola. 

 

Option: 1

\frac{5 a^2}{4}


Option: 2

\frac{15 a^2}{4}


Option: 3

\frac{5 a^2}{2}


Option: 4

\frac{15 a^2}{2}


Answers (1)

best_answer

Equation of tangent at \mathrm{\mathrm{P}\left(\mathrm{at}^2, 2 \mathrm{at}\right) }
is \mathrm{\quad t y=x+a t^2 \Rightarrow x-t y+a t^2=0 }........................(1)
which is also tangent to the circle
\mathrm{\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2 / 2 }..........................................(2)
then length of perpendicular from center of (2) to (1) = radius of the circle
\mathrm{\therefore \quad\left|\frac{a t^2}{\sqrt{1+t^2}}\right|=\frac{a}{\sqrt{2}} }

or
\mathrm{2 t^4=\left(1+t^2\right) }
or
\mathrm{\left(\mathrm{t}^2-1\right)\left(2 \mathrm{t}^2+1\right)=0 }
\mathrm{\therefore \quad 2 \mathrm{t}^2+1 \neq 0 \quad \therefore \quad \mathrm{t}^2-1=0 }
then \mathrm{\mathrm{t}= \pm 1 }
then co-ordinates of P and Q are (a, 2 a) and (a,-2 a) respectively.
\mathrm{\therefore \quad P Q=4 a }
\mathrm{\therefore } Equation of tangent at \mathrm{\mathrm{P}(\mathrm{a}, 2 \mathrm{a}) \quad } is \mathrm{\mathrm{x}-\mathrm{y}+\mathrm{a}=0 }................(3)
\mathrm{\{from \, \, equation (1)\} }
Let R be \mathrm{\left(x_1, y_1\right) }
Then equation of tangent at \mathrm{R\left(x_1, y_1\right) on (2) } is
\mathrm{\mathrm{xx}_1+\mathrm{yy}_1=\frac{\mathrm{a}^2}{2} }.........................................................(4)
Hence (3) and (4) are identical
\mathrm{\begin{array}{ll} \therefore & \frac{\mathrm{x}_1}{1}=\frac{\mathrm{y}_1}{-1}=-\frac{\mathrm{a}}{2} \\ \therefore & \left(\mathrm{x}_1, \mathrm{y}_1\right)=\left(-\frac{\mathrm{a}}{2}, \frac{\mathrm{a}}{2}\right) \end{array} }
Hence co-ordinate of\mathrm{ \mathrm{S}^{\prime} is \left(-\frac{\mathrm{a}}{2},-\frac{\mathrm{a}}{2}\right) } 
\therefore \quad \mathrm{RS}^{\prime}=\mathrm{a}
Since Quadrilateral \mathrm{PQRS} is trapezium whose area
\mathrm{\begin{aligned} & =\frac{1}{2}\left(P Q+R S^{\prime}\right) \times\left(a+\frac{a}{2}\right) \\ & =\frac{1}{2} \times(4 a+a) \times \frac{3 a}{2} \end{aligned} }
\mathrm{=\frac{15 \mathrm{a}^2}{4} \text { sq. units. }}

Posted by

HARSH KANKARIA

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