Final Matching (List–I → List–II)

Answer: (A) → (IV), (B) → (I), (C) → (II), (D) → (III)

Solutions (with steps)

(A) \(x^2-4x+4y+4y^2=12\) \[ (x-2)^2-4+4\!\left[(y+\tfrac12)^2-\tfrac14\right]=12 \;\Rightarrow\; (x-2)^2+4(y+\tfrac12)^2=17 \] \[ \frac{(x-2)^2}{17}+\frac{(y+\tfrac12)^2}{17/4}=1 \] Ellipse with \(a^2=17,\; b^2=\tfrac{17}{4}\Rightarrow e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\tfrac14}=\frac{\sqrt{3}}{2}. \)

(B) \(5x^2+9y^2=45 \Rightarrow \frac{x^2}{9}+\frac{y^2}{5}=1\). Here \(a=\sqrt{9}=3,\; b=\sqrt{5}\). Latus rectum length (ellipse) \(= \dfrac{2b^2}{a}=\dfrac{2\cdot5}{3}=\dfrac{10}{3}.\)

(C) Tangent: \(x+y=a \Rightarrow y=-x+a\). Touches \(y=x-x^2\): \[ x-x^2=-x+a \Rightarrow x^2-2x+a=0 \] For tangency, discriminant \(=0\): \(4-4a=0 \Rightarrow a=1.\)

(D) \(3x^2-y^2=4 \Rightarrow \dfrac{x^2}{4/3}-\dfrac{y^2}{4}=1\). Hyperbola with \(a^2=\tfrac{4}{3},\, b^2=4\). \[ e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{4}{4/3}}=\sqrt{1+3}=2. \]