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Parabola and Circle Tangents: Area of Triangle PRS and lambda^2 | JEE

JEE Maths question with a full step-by-step solution.

Question
Consider the curves
C1:∣z−2∣=2+Re⁡(z)andC2:∣z∣=3,C_1 : |z-2| = 2 + \operatorname{Re}(z) \qquad \text{and} \qquad C_2 : |z| = 3 ,
 where z=x+iyz = x + iy, x,y∈Rx, y \in \mathbb{R} and i=−1i = \sqrt{-1}. They intersect at PP and QQ in the first and fourth quadrants respectively. Tangents to C1C_1 at PP and QQ intersect the xx-axis at RR, and tangents to C2C_2 at PP and QQ intersect the xx-axis at SS. If the area of △PRS\triangle PRS is λ2\lambda\sqrt2 sq. units, then find (λ2)\left(\lambda^{2}\right).
Solution
Answer: 100 (± 0.01)
Step 1: Convert C1C_1 to Cartesian form. With z=x+iyz = x+iy, Re⁡(z)=x\operatorname{Re}(z) = x:
(x−2)2+y2=x+2.\sqrt{(x-2)^{2} + y^{2}} = x + 2 .
 Step 2: Square both sides and expand.
(x−2)2+y2=(x+2)2  ⟹  y2=(x+2)2−(x−2)2.(x-2)^{2} + y^{2} = (x+2)^{2} \implies y^{2} = (x+2)^{2} - (x-2)^{2} .
 Step 3: Use A2−B2=(A+B)(A−B)A^{2}-B^{2} = (A+B)(A-B) with A=x+2A = x+2, B=x−2B = x-2:
y2=(2x)(4)=8x,y^{2} = (2x)(4) = 8x ,
 so C1C_1 is the parabola y2=8xy^{2} = 8x. Meanwhile C2C_2 is the circle x2+y2=9x^{2}+y^{2} = 9. Question attachment Step 4: Find the intersection points. Substituting y2=8xy^{2} = 8x into the circle,
x2+8x=9  ⟹  x2+8x−9=0  ⟹  (x+9)(x−1)=0.x^{2} + 8x = 9 \implies x^{2}+8x-9 = 0 \implies (x+9)(x-1) = 0 .
 Only x=1x = 1 is admissible (x=−9x = -9 gives y2<0y^{2} < 0). Then y2=8y^{2} = 8, so
P(1, 22),Q(1, −22).P\left(1,\ 2\sqrt2\right), \qquad Q\left(1,\ -2\sqrt2\right).
 Step 5: Find RR using the tangent to the parabola. For y2=4axy^{2} = 4ax with 4a=84a = 8, the tangent at (x1,y1)\left(x_1,y_1\right) is yy1=4(x+x1)yy_1 = 4\left(x + x_1\right). At PP:
y(22)=4(x+1).y\left(2\sqrt2\right) = 4(x+1).
 Step 6: Put y=0y = 0 to meet the xx-axis.
0=4(x+1)  ⟹  x=−1  ⟹  R(−1,0).0 = 4(x+1) \implies x = -1 \implies R(-1, 0).
 (By symmetry the tangent at QQ meets the axis at the same point, so RR is well defined.) Step 7: Find SS using the tangent to the circle. For x2+y2=9x^{2}+y^{2} = 9 the tangent at (x1,y1)\left(x_1,y_1\right) is xx1+yy1=9xx_1 + yy_1 = 9. At PP:
x(1)+y(22)=9.x(1) + y\left(2\sqrt2\right) = 9 .
 Step 8: Put y=0y = 0.
x=9  ⟹  S(9,0).x = 9 \implies S(9, 0).
 Step 9: Compute the area of △PRS\triangle PRS. Both RR and SS lie on the xx-axis, so take RSRS as the base; the height is the vertical distance of PP from that axis.
RS=9−(−1)=10,height=22.RS = 9 - (-1) = 10, \qquad \text{height} = 2\sqrt2 .
Area=12×10×22=102.\text{Area} = \frac12 \times 10 \times 2\sqrt2 = 10\sqrt2 .
 Step 10: Compare with λ2\lambda\sqrt2.
λ2=102  ⟹  λ=10  ⟹  λ2=100.\lambda\sqrt2 = 10\sqrt2 \implies \lambda = 10 \implies \lambda^{2} = 100 .
 Answer: 100100 (i.e. 100.00100.00).
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