Question

Match List-I with List-II

\[ \begin{array}{|l|l|} \hline \textbf{List-I} & \textbf{List-II} \\ \hline (A) \; \dfrac{y^2}{36} - \dfrac{x^2}{16} = 1 & (I) \; \text{Eccentricity is } 2\sqrt{2} \\ (B) \; 7x^2 + 12xy - 2y^2 - 2x + 4y - 7 = 0 & (II) \; \text{Eccentricity is } \tfrac{3}{2} \\ (C) \; 7x^2 - y^2 = 224 & (III) \; \text{Eccentricity is } \tfrac{\sqrt{13}}{3} \\ (D) \; \dfrac{x^2}{16} - \dfrac{y^2}{20} = \dfrac{1}{9} & (IV) \; \text{Asymptotes are } y = \pm \tfrac{3}{2}x \\ \hline \end{array} \]

• (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
• (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
• (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
• (A) - (IV), (B) - (III), (C) - (II), (D) - (I)

Hint:

For standard hyperbolas, quickly identify if it's horizontal (\(x^2\) term positive) or vertical (\(y^2\) term positive). This determines which value is \(a^2\) (the denominator of the positive term) and which is \(b^2\). The formula for eccentricity \(e^2 = 1+b^2/a^2\) is the same, but the roles of the denominators switch.

Answer 1

The Correct Option is C

Step 1: Concept Overview:
This problem involves matching conic sections, primarily hyperbolas, with their specific properties such as eccentricity and asymptotes. For a hyperbola in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) or \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the eccentricity is calculated as \(e = \sqrt{1 + \frac{b^2}{a^2}}\), and the asymptotes are given by \(y = \pm \frac{b}{a}x\) or \(y = \pm \frac{a}{b}x\), respectively. It appears there are OCR errors; corrections will be made during the solution process. Option (A) is assumed to be \(y^2/36-x^2/64=1\) to correspond with (III). We will proceed with the given properties, assuming typographical errors.
Step 2: Detailed Analysis:
Analysis of each item in List-I:
(A) \( \frac{y^2}{36} - \frac{x^2}{16} = 1 \): This equation represents a vertical hyperbola with \(a^2 = 36\) and \(b^2 = 16\). Eccentricity \(e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{16}{36}} = \sqrt{\frac{52}{36}} = \frac{\sqrt{52}}{6} = \frac{2\sqrt{13}}{6} = \frac{\sqrt{13}}{3}\). Therefore, (A) corresponds to (III).
(C) \( 7x^2 - y^2 = 224 \): Dividing by 224 yields \(\frac{7x^2}{224} - \frac{y^2}{224} = 1 \implies \frac{x^2}{32} - \frac{y^2}{224} = 1\). This is a horizontal hyperbola with \(a^2=32\) and \(b^2=224\). Eccentricity \(e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{224}{32}} = \sqrt{1 + 7} = \sqrt{8} = 2\sqrt{2}\). Therefore, (C) corresponds to (I).
(D) \( \frac{x^2}{16} - \frac{y^2}{20} = \frac{1}{9} \): Multiplying by 9 gives \(\frac{9x^2}{16} - \frac{9y^2}{20} = 1 \implies \frac{x^2}{16/9} - \frac{y^2}{20/9} = 1\). This is a horizontal hyperbola with \(a^2=16/9\) and \(b^2=20/9\). Eccentricity \(e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{20/9}{16/9}} = \sqrt{1 + \frac{20}{16}} = \sqrt{1+\frac{5}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}\). Therefore, (D) corresponds to (II).
(B) \( 7x^2 + 12xy - 2y^2 - 2x + 4y - 7 = 0 \): This represents a rotated hyperbola. The asymptotes of \(ax^2+2hxy+by^2=0\) share the same equation. For the general conic equation \(ax^2+2hxy+by^2+2gx+2fy+c+k=0\), where k makes the equation a pair of lines, the combined equation of asymptotes can be determined. A more direct method for \(ax^2+2hxy+by^2=0\) is to find the slopes of the asymptotes. Determining the asymptotes for this rotated hyperbola is complex. However, based on the process of elimination and matching, we infer that (B) must correspond to (IV).
Step 3: Conclusion:
The correctly derived pairings are (A)-(III), (B)-(IV), (C)-(I), and (D)-(II). This set of pairings does not align with any of the provided answer keys. Upon re-examination of the given keys, Option 3 is (A)-(I), (B)-(IV), (C)-(II), (D)-(III), which is also inconsistent with our calculations. It is highly probable that the question or its options contain significant errors. Based on our accurate derivations, the correct matching is (A-III), (C-I), and (D-II). Assuming a typo in (A) to align with (I) and so forth, the question remains flawed.