Question

Let S be the focus of the parabola \[{y^2}{\text{ }} = {\text{ }}8x\] and let PQ be the common chord of the circle \[{x^2}{\text{ }} + {\text{ }}{y^2}{\text{ }} - 2x{\text{ }} - 4y{\text{ }} = 0\] and the given parabola. The area of the \[\vartriangle PQS\] is
A. 4
B. 5
C. 6
D. 7

Answer 1

Hint: PARABOLA :
A parabola is the locus of a point which moves in a plane, such that its distance from a fixed point (focus) is equal to its perpendicular distance from a fixed straight line (directrix).
Standard equation of a parabola is ${y^2} = 4ax$
 For this parabola :
Vertex is (0, 0)
Focus is (a, 0)
Axis is y = 0
Directrix is x + a = 0
Using the equation of circle find the general equation of circle. Now assuming (x, y) as \[(2{t^2},4t)\] in this equation we will simplify the equation to find real roots for t. From this we will find Points P and Q which are touching the parabola and the circle. From the parabola equation we will find the focus S coordinates. Now with the help of distance formula we will find the value of altitude QS and base PS. As we now know the value of altitude and base we will find the area of \[\vartriangle PQS\] .

Complete step-by-step solution:
Given Parabola: \[{y^2}{\text{ }} = {\text{ }}8x\]
Given Circle:
 \[{x^2}{\text{ }} + {\text{ }}{y^2}{\text{ }} - 2x{\text{ }} - 4y{\text{ }} = 0\]
Reducing the equation as general equation of circle:
 \[ \Rightarrow {(x - 1)^2} + {(y - 2)^2} = 5\]
Take a point on the parabola as \[(2{t^2},4t)\]
Solve equations simultaneously
 \[ \Rightarrow {(2{t^2} - 1)^2} + {(4t - 2)^2} = 5\]
Opening the brackets and using the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$ for expansion of square
 \[ \Rightarrow 4{t^4} - 4{t^2} + 1 + 16{t^2} - 16t + 4 = 5\]
Simplifying further, we get
 \[
   \Rightarrow 4{t^4} + 12{t^2} - 16t + 5 = 5 \\
   \Rightarrow {t^4} + {t^2} - 4t = 0 \\
 \]
Taking t common and factorizing the remaining equation, we get
 \[
   \Rightarrow t(t - 1)({t^2} + t + 4) = 0 \\
   \Rightarrow t = 0,1,not\;real \\
 \]
So, $t = 0,1$are the only real numbers as \[({t^2} + t + 4)\] does not have real roots.
By putting the values t=0 and 1 of coordinates in the \[(2{t^2},4t)\]
Hence, for common chord endpoints, we get \[P(0,0)\] and \[Q(2,4)\] .
For t=1, point Q $({x_2},{y_2}) = (2,4)$
For t=0, point P $({x_3},{y_3}) = (0,0)$
The given parabola \[{y^2}{\text{ }} = {\text{ }}8x\] is of the form ${y^2} = 4ax$where, $4a = 8,a = 2$.
The coordinates of focus are$(a,0)\;i.e.S(2,0)$.
 \[\vartriangle PQS\] will form a right angled triangle with area using distance formula
Where S(2,0) Q(2,4) and P (0,0)
 \[
  Area = \dfrac{1}{2} \times SQ \times PS \\
   \Rightarrow Area = \dfrac{1}{2} \times \sqrt {{{\left( {2 - 2} \right)}^2} + {{(4 - 0)}^2}} \times \sqrt {{{\left( {2 - 0} \right)}^2} + {{(0 - 0)}^2}} \\
   \Rightarrow Area = \dfrac{1}{2} \times \sqrt {16} \times \sqrt 4 \\
   \Rightarrow Area = \dfrac{1}{2} \times 4 \times 2 = 4 \\
 \]
The area of the \[\vartriangle PQS\] is 4 square units

So, option (A) is the correct answer.

Note: The discriminant is a value calculated from a quadratic equation. It uses it to 'discriminate' between the roots (or solutions) of a quadratic equation. A quadratic equation is one of the form : \[a{x^{_2}}{\text{ }} + {\text{ }}bx{\text{ }} + {\text{ }}c\] . The discriminant, \[D{\text{ }} = {\text{ }}{b^2}{\text{ }} - {\text{ }}4ac\] , have three cases:
If the discriminant is greater than zero, the quadratic equation has two real, distinct (different) roots.
If the discriminant is less than zero, the quadratic equation has no real roots.
If the discriminant is equal to zero, the quadratic equation has two real, identical roots.
In this question \[({t^2} + t + 4)\] discriminant is -15 which is less than zero. Therefore this equation doesn’t have any real roots.