Question 46

Two pairs of straight lines $$x^2 — 7x + 6 = 0$$ and $$y^2 — 14y + 40 = 0$$ intersect to form a rectangle. Let the diagonals of the rectangle intersect at the point W. A circle with center W and with tangents as lines $$y^2 — 14y + 40 = 0$$ intersects lines $$x^2 — 7x + 6 = 0$$ at points P, Q, R, S. The area of the rectangle PQRS is

Solution

$$x^2 — 7x + 6 = 0$$ => x= 1, x=6$$PM\ =\sqrt{\ PW^2-MW^2}$$

$$y^2 — 14y + 40 = 0$$ => y= 4, y=10

$$W\ =\ \left(\frac{1+6}{2},\frac{4+10}{2}\right)\ =\ \left(\frac{7}{2},7\right)$$

QR = 6-1 =5

r = (10-4)/2 = 3

PW = r= 3

MW = QR/2 = 5/2

$$PM\ =\sqrt{\ PW^2-MW^2}$$ = $$\frac{\sqrt{\ 11}}{2}$$

PQ=2PM = $$\sqrt{\ 11}$$

Area of PQRS = QR x PQ = $$5\sqrt{\ 11}$$

Share on Whatsapp Share on Whatsapp

Create a FREE account and get: