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Solution


Q.

Let C be the circle x2+(y–1)2=2, E1 and E2 be two ellipses whose centres lie at the origin and major axes lie on x-axis and y-axis respectively. Let the straight line x + y = 3 touch the curves C, E1 and E2 at P(x1,y1), Q(x2,y2) and R(x3,y3) respectively. Given that P is the mid-point of the line segment QR and PQ=223, the value of 9(x1y1+x2y2+x3y3) is equal to __________.          [2025]

Ans.

(46)

(a) Solving the line x + y = 3, and the circle x2+(y–1)2=2

Substitute y = 3 – x

x2+(3–x–1)2=2

⇒ x2–2x+1=0

⇒ x=1 ⇒ y=2

So, P=(x1,y1)=(1,2) ⇒ x1y1=2

Use mid-point condition

Let Q=(x2,y2), R=(x3,y3).

Since P is the mid-point of QR

⇒ x2+x3=2x1=2, y2+y3=2y1=4

So, we can write : x3=2–x2, y3=4–y2

(b) Given

PQ=223 ⇒ PQ2=(x2–1)2+(y2–2)2=89

Let's denote : x2=a, y2=b, x3=2–a, y3=5–b

(a–1)2+(b–2)2=89

⇒ a2–2a+1+b2–4b+4=89

⇒ a2+b2–2a–4b+5=89

⇒ 9a2+9b2–18a–36b+37=0

Hence, a=53, b=43

x1y1+x2y2+x3y3=2+ab+(2–a)(4–b)

9(x1y1+x2y2+x3y3)=9(10+2ab–2b–4a)

= 90 + 18ab – 18b – 36a = 46.