Information Given:
A children’s park has a circular garden as shown in the figure with DC and AB representing x and y-axis respectively. AB and CD are two cemented walkways of length 70 m each. Areas marked as AEOF and BGOH are flower beds whereas ECGO and FDHO are covered with grass. FOE and HOG are parabolas with focus (0, 6) and (0, −6) respectively.
Based on the information given above answer the questions.
There is a plan to connect EG and FH through a straight mud path. What will be the equation of both?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(x=10\sqrt{6}\) and \(x=-10\sqrt{6}\)
The garden is the circle \(x^2+y^2 = 1225\) (radius \(\frac{70}{2} = 35\)). The parabola FOE (focus (0, 6), vertex O) is \(x^2 = 24y\), and HOG (focus (0, −6)) is \(x^2 = -24y\).
E and F are the points where FOE meets the circle. Putting \(x^2 = 24y\) in the circle:
\(24y+y^2 = 1225\)
\(y^2+24y-1225 = 0\)
\((y-25)(y+49) = 0\)
Since \(y = \frac{x^2}{24} \ge 0\), y = 25, and then
\(x^2 = 24 \times 25 = 600\)
\(x = \pm 10\sqrt{6}\)
So \(E(10\sqrt{6}, 25)\) and \(F(-10\sqrt{6}, 25)\). In the same way HOG meets the circle at y = −25, giving \(G(10\sqrt{6}, -25)\) and \(H(-10\sqrt{6}, -25)\).
E and G have the same x-coordinate, and so do F and H. Hence
\(EG: x = 10\sqrt{6}\)
\(FH: x = -10\sqrt{6}\)