Let C be the circle $x^2 + (y-1)^2 = 2, E_1$ and $E_2$ 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, $E_1$ and $E_2$ at P($x_1,y_1$), Q($x_2,y_2$) and R($x_3,y_3$) respectively. Given that P is the midpoint of the line segment QR and PQ = $\frac{2\sqrt{2}}{3}$ the value of 9($x_1y_1 + x_2y_2 + x_3y_3$) is equal to _____

Step 1: Find the coordinates of point P

The line $x+y=3$ touches the circle $x^2+(y-1)^2=2$ at point $P(x_1,y_1)$. 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 $P(x_1,y_1)$ is $\frac{y_1-1}{x_1-0}$. 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, $\frac{dx}{dy} = -\frac{x}{y-1}$, and set $\frac{dy}{dx} = -1$. Solving the system of equations $x_1+y_1=3$ and $x_1=y_1-1$ 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 $E_1$ and $E_2$ at $Q(x_2,y_2)$ and $R(x_3,y_3)$ respectively. We are given that $P$ is the midpoint of the line segment $QR$ and $PQ = \frac{2\sqrt{2}}{3}$. 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 $\sqrt{(1+d-1)^2 + (2-d-2)^2} = \sqrt{d^2 + (-d)^2} = \sqrt{2d^2} = |d|\sqrt{2}$. Equating this to the given distance $PQ = \frac{2\sqrt{2}}{3}$, we find $|d| = \frac{2}{3}$. We can choose $d = \frac{2}{3}$ (the choice of sign for $d$ just swaps $Q$ and $R$). This gives $Q(\frac{5}{3}, \frac{4}{3})$ and $R(\frac{1}{3}, \frac{8}{3})$.

Step 3: Identify which point corresponds to which ellipse

Ellipse $E_1$ 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 $E_2$ has its major axis on the $y$ -axis, meaning its $y$ -radius is greater than its $x$ -radius ($b>a$). For $E_2$, the tangent point $(x,y)$ will generally have $|y| > |x|$ (unless $x=0$). Comparing the coordinates of $Q(\frac{5}{3}, \frac{4}{3})$ and $R(\frac{1}{3}, \frac{8}{3})$: for $Q$, $|x_Q| = \frac{5}{3}$ and $|y_Q| = \frac{4}{3}$, so $|x_Q| > |y_Q|$. For $R$, $|x_R| = \frac{1}{3}$ and $|y_R| = \frac{8}{3}$, so $|y_R| > |x_R|$. Thus, $Q$ is the tangency point for $E_1$ and $R$ is the tangency point for $E_2$.

Step 4: Calculate the required expression

Now we have all the coordinates: $P(x_1,y_1) = (1{,}2)$, $Q(x_2,y_2) = (\frac{5}{3}, \frac{4}{3})$, and $R(x_3,y_3) = (\frac{1}{3}, \frac{8}{3})$. We need to calculate $9(x_1y_1 + x_2y_2 + x_3y_3)$. First, calculate the product $x_iy_i$ for each point. For $P$, $x_1y_1 = 1 \times 2 = 2$. For $Q$, $x_2y_2 = \frac{5}{3} \times \frac{4}{3} = \frac{20}{9}$. For $R$, $x_3y_3 = \frac{1}{3} \times \frac{8}{3} = \frac{8}{9}$. Summing these products gives $2 + \frac{20}{9} + \frac{8}{9} = 2 + \frac{28}{9} = \frac{18+28}{9} = \frac{46}{9}$. Finally, multiply by 9: $9 \times \frac{46}{9} = 46$.

More from this chapter

• Find the equation of the circle passing through points (1,2), (3,4), and (5,2).
• Determine the equation of the circle with radius 4 and Centre (-2, 3).
• Find the equation of the hyperbola where foci are (0, +-6) and the length of the latus rectum is 36.
• Find the equation of the hyperbola where foci are (0, ± 12) and the length of the latus rectum is 36
• Find the equation of the circle with centre (-3,2) and radius 10.
• Find the equation of the circle with centre (-3,2) and radius 9.
• Find the equation of the circle with centre (-3,2) and radius 8.
• Find the equation of the circle with centre (-3,2) and radius 6.

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