Exercise 6.5 Class 12 Math Solutions - Unit 6 Conics, Punjab Board

Step-by-step solutions to Exercise 6.5, Unit 6: Conics, for Class 12 Punjab Board students. Covers the definition of a hyperbola with eccentricity e > 1, derivation of standard equation x²/a² - y²/b² = 1, key features (foci (±c, 0), transverse and conjugate axes, vertices, directrices x = ±a/e, asymptotes y = ±(b/a)x, latus rectum 2b²/a), rectangular hyperbola (a = b), translated forms, and equations of tangents and normals including the tangency condition c² = a²m² - b², worked out question by question.

This page provides complete step-by-step solved solutions for Exercise 6.5 of Unit 6: Conics, Class 12 Mathematics, for Punjab Board (PCTB / PECTAA). All textbook problems are solved with comprehensive analytical rigor, covering the definition of a hyperbola with eccentricity \(e > 1\), derivation of standard forms \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) (horizontal) and \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) (vertical), fundamental relation \(c^2 = a^2 + b^2\), key elements including foci \((\pm c, 0)\), vertices \((\pm a, 0)\), directrices \(x = \pm \frac{a}{e}\), asymptotes \(y = \pm \frac{b}{a}x\), eccentricity \(e = \frac{c}{a} = \frac{\sqrt{a^2 + b^2}}{a} > 1\), length of latus rectum \(\frac{2b^2}{a}\), rectangular hyperbola (\(a = b\)), translated central forms \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), equation of tangent \(\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1\), tangency condition \(c^2 = a^2 m^2 - b^2\), and equations of normals.

12
Mathematics
Exercise 6.5

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Exercise 6.4

Exercise 6.6

Exercise Overview

Exercise 6.5 of Unit 6: Conics in Class 12 Mathematics under the Punjab Board PCTB / PECTAA curriculum focuses on the hyperbola, defined as the conic section with eccentricity (e > 1), where the absolute difference of distances from two fixed foci (F) and (F') to any point on the curve is a positive constant ((||PF| - |PF'|| = 2a)). The standard central equation of a horizontal hyperbola is (rac{x^2}{a^2} - rac{y^2}{b^2} = 1) (where (c^2 = a^2 + b^2)), featuring centre at (C(0, 0)), foci at ((pm c, 0)), transverse axis of length (2a) along the (x)-axis with vertices ((pm a, 0)), conjugate axis of length (2b) along the (y)-axis, directrices (x = pm rac{a}{e}), eccentricity (e = rac{c}{a} = rac{sqrt{a^2 + b^2}}{a} > 1), asymptotes (y = pm rac{b}{a}x), and latus rectum length (rac{2b^2}{a}). For a vertical hyperbola (rac{y^2}{a^2} - rac{x^2}{b^2} = 1), the transverse axis lies along the (y)-axis with foci ((0, pm c)), vertices ((0, pm a)), directrices (y = pm rac{a}{e}), and asymptotes (y = pm rac{a}{b}x).

The exercise explores rectangular hyperbolas (where (a = b), giving (x^2 - y^2 = a^2) with eccentricity (e = sqrt{2}) and perpendicular asymptotes (y = pm x)) and establishes translated hyperbola theorems: Theorem 1 covers translated horizontal hyperbolas (rac{(x - h)^2}{a^2} - rac{(y - k)^2}{b^2} = 1) having centre (C(h, k)), foci ((h pm c, k)), vertices ((h pm a, k)), and asymptotes (y - k = pm rac{b}{a}(x - h)); Theorem 2 covers translated vertical hyperbolas (rac{(y - k)^2}{a^2} - rac{(x - h)^2}{b^2} = 1). Furthermore, students derive the equation of the tangent at (P(x_1, y_1)) as (rac{xx_1}{a^2} - rac{yy_1}{b^2} = 1), the condition of tangency for (y = mx + c) as (c^2 = a^2 m^2 - b^2) giving slope form (y = mx pm sqrt{a^2 m^2 - b^2}) with contact points (left(mp rac{a^2 m}{c}, mp rac{b^2}{c} ight)), and the equation of the normal line.

For Punjab Board HSSC Part 2 examinations across all nine educational divisions (Lahore, Rawalpindi, Gujranwala, Faisalabad, Multan, Bahawalpur, Sargodha, Sahiwal, and D.G. Khan), Exercise 6.5 concludes Unit 6 and provides critical Section B Short Questions (extracting foci, vertices, directrices, eccentricity, asymptotes, and latus rectum; formulating hyperbola equations from given features) and high-scoring Section C Long Questions (completing squares for general second-degree equations (Ax^2 - By^2 + Gx + Fy + C = 0), deriving tangent/normal equations, and proving geometric intersection properties). These step-by-step solutions provide complete algebraic working, asymptote derivations, and full coordinate proofs for every textbook question.

FAQs

Are these Exercise 6.5 solutions according to the Punjab Board syllabus?

Yes. These solutions follow the Punjab Curriculum and Textbook Board (PCTB / PECTAA) syllabus for Class 12 Mathematics, matching all hyperbola concepts and problems in Unit 6, Exercise 6.5 for the 2026-2027 session.

Do these solutions cover every question in Exercise 6.5?

Yes, every question in Exercise 6.5 is completely solved, including finding hyperbola elements (foci, vertices, directrices, asymptotes, eccentricity, latus rectum), constructing hyperbola equations from given conditions, rectangular hyperbola cases, translated equations, and tangent lines.

Are the solutions step-by-step?

Yes. Each problem details the exact analytical steps-identifying horizontal vs vertical hyperbola orientations, using the fundamental relation c² = a² + b², calculating eccentricity e = c/a > 1, finding asymptotes y = ±(b/a)x, and applying condition c² = a²m² - b².

Can students read these solutions online?

Yes, the complete Exercise 6.5 solution set is available to read online directly on this page on MaryamNotes.pk with zero software download or registration required.

Are these solutions based on the latest PCTB / PECTAA syllabus?

Yes, they are strictly aligned with the updated PCTB and PECTAA Mathematics curriculum for Class 12 HSSC Part 2 for the 2026-2027 academic year.

Are these solutions free for students?

Yes, Exercise 6.5 solutions on this page are completely free for all Intermediate Part 2 students across Lahore, Rawalpindi, Gujranwala, Faisalabad, Multan, and all other Punjab boards.