Area Ques 9
- The common tangents to the circle $x^2+y^2=2$ and the parabola $y^2=8 x$ touch the circle at the points $P, Q$ and the parabola at the points $R, S$. Then, the area (in $s q$ units) of the quadrilateral $P Q R S$ is
(2014 Adv.)
(a) $3$
(b) $6$
(c) $9$
(d) $15$
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Answer:
Correct Answer: 9.(d)
Solution: (d) PLAN
(i) $y=m x+a / m$ is an equation of tangent to the parabola $y^2=4 a x$.
(ii) A line is a tangent to circle, if distance of line from centre is equal to the radius of circle.
(iii) Equation of chord drawn from exterior point $\left(x_1, y_1\right)$ to a circle/parabola is given by $T=0$.
(iv) Area of trapezium $=\frac{1}{2}$ (Sum of parallel sides)
Let equation of tangent to parabola be $y=m x+\frac{2}{m}$
It also touches the circle $x^2+y^2=2$.
$\therefore \quad \left|\frac{2}{m \sqrt{1+m^2}}\right|=\sqrt{2}$
$\Rightarrow \quad m^4+m^2=2 \Rightarrow m^4+m^2-2=0$
$\Rightarrow \quad\left(m^2-1\right)\left(m^2+2\right)=0$
$\Rightarrow \quad m= \pm 1, m^2=-2 \quad$ [rejected $\left.m^2=-2\right]$
So, tangents are $y=x+2, y=-x-2$.
They, intersect at $(-2,0)$.
Equation of chord $P Q$ is $-2 x=2 \Rightarrow x=-1$
Equation of chord $R S$ is $O=4(x-2) \Rightarrow x=2$
$\therefore \quad$ Coordinates of $P, Q, R, S$ are
$P(-1,1), Q(-1,-1), R(2,4), S(2,-4)$
$\therefore \quad$ Area of quadrilateral $=\frac{(2+8) \times 3}{2}=15$ sq units
BETA
