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 the 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 midpoint of the line segment QR and PQ = 322 the value of 9(x1y1+x2y2+x3y3) is equal to _____
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Step-by-Step Solution
Step 1: Find the coordinates of point P
The line x+y=3 touches the circle x2+(y−1)2=2 at point P(x1,y1). We can find the coordinates of P by using the condition that the tangent line is perpendicular to the radius at the point of tangency. The slope of the tangent line x+y=3 is −1. The slope of the radius from the center (0,1) to P(x1,y1) is x1−0y1−1. Since the tangent is perpendicular to the radius, the product of their slopes is −1. Alternatively, we can find the derivative of the circle's equation, dydx=−y−1x, and set dxdy=−1. Solving the system of equations x1+y1=3 and x1=y1−1 along with the circle equation gives two possible points for P: (1,2) and (−1,0). Since the line x+y=3 has a positive y -intercept, the point (1,2) is the correct tangency point.
Step 2: Find the coordinates of Q and R
We found P(1,2). The line x+y=3 touches the ellipses E1 and E2 at Q(x2,y2) and R(x3,y3) respectively. We are given that P is the midpoint of the line segment QR and PQ=322. Since P is the midpoint, PR must also be equal to PQ. The points Q and R lie on the line x+y=3. We can parameterize points on this line relative to P. Let Q=(1+d,2−d) and R=(1−d,2+d). The distance PQ is (1+d−1)2+(2−d−2)2=d2+(−d)2=2d2=∣d∣2. Equating this to the given distance PQ=322, we find ∣d∣=32. We can choose d=32 (the choice of sign for d just swaps Q and R). This gives Q(35,34) and R(31,38).
Step 3: Identify which point corresponds to which ellipse
Ellipse E1 has its major axis on the x -axis, meaning its x -radius is greater than its y -radius (a>b). For an ellipse centered at the origin, if the major axis is along the x -axis, the tangent point (x,y) will generally have |x| > |y| (unless y=0). Ellipse E2 has its major axis on the y -axis, meaning its y -radius is greater than its x -radius (b>a). For E2, the tangent point (x,y) will generally have |y| > |x| (unless x=0). Comparing the coordinates of Q(35,34) and R(31,38): for Q, ∣xQ∣=35 and ∣yQ∣=34, so ∣xQ∣>∣yQ∣. For R, ∣xR∣=31 and ∣yR∣=38, so ∣yR∣>∣xR∣. Thus, Q is the tangency point for E1 and R is the tangency point for E2.
Step 4: Calculate the required expression
Now we have all the coordinates: P(x1,y1)=(1,2), Q(x2,y2)=(35,34), and R(x3,y3)=(31,38). We need to calculate 9(x1y1+x2y2+x3y3). First, calculate the product xiyi for each point. For P, x1y1=1×2=2. For Q, x2y2=35×34=920. For R, x3y3=31×38=98. Summing these products gives 2+920+98=2+928=918+28=946. Finally, multiply by 9: 9×946=46.