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Karnataka PUC I Straight Lines and Conic Sections — practice questions

Explore 50 free Straight Lines and Conic Sections questions in Karnataka PUC I Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Straight Lines and Conic Sections questions and exercises with answers

Showing questions 1–30 of 50.

Equation of circle with center (h, k) and radius r:Read the answer and worked solution →Standard ellipse x²/a² + y²/b² = 1 (a > b) has eccentricity:Read the answer and worked solution →Hyperbola x²/a² - y²/b² = 1 has eccentricity:Read the answer and worked solution →Latus rectum of parabola y² = 4ax:Read the answer and worked solution →Equation of tangent to parabola y² = 4ax at point (at², 2at):Read the answer and worked solution →Axis of symmetry of parabola x² = 4ay (a > 0):Read the answer and worked solution →Asymptotes of hyperbola x²/a² - y²/b² = 1:Read the answer and worked solution →Equation of normal to parabola y² = 4ax at (at², 2at):Read the answer and worked solution →Conjugate hyperbola of x²/a² - y²/b² = 1:Read the answer and worked solution →Equation of circle passing through origin, (a, 0), (0, b):Read the answer and worked solution →Tangent to circle x² + y² = r² at point (x₁, y₁):Read the answer and worked solution →Find focus and directrix of parabola y² = -8x:Read the answer and worked solution →Eccentricity of hyperbola 9x² - 16y² = 144:Read the answer and worked solution →Length of latus rectum of ellipse x²/25 + y²/9 = 1:Read the answer and worked solution →Common tangent to circle x² + y² = 5 and parabola y² = 8x:Read the answer and worked solution →Eccentricity of ellipse: x² + 4y² = 4:Read the answer and worked solution →For parabola y² = 4ax, point (at², 2at) gives parametric form. If t = 2 and a = 3:Read the answer and worked solution →Sum of distances from any point on ellipse 9x² + 16y² = 144 to its foci:Read the answer and worked solution →Difference of focal distances from any point on hyperbola x²/16 - y²/9 = 1:Read the answer and worked solution →For hyperbola x²/25 - y²/144 = 1, find length of conjugate axis:Read the answer and worked solution →Auxiliary circle of ellipse x²/a² + y²/b² = 1:Read the answer and worked solution →Relation between eccentricities of conjugate hyperbolas:Read the answer and worked solution →If the line $x - 1 = 0$ is the directrix of the parabola $y^2 - kx + 8 = 0$, then one of the values of $k$ isRead the answer and worked solution →The sum of the focal distances from any point on the ellipse $9x^2 + 16y^2 = 144$ isRead the answer and worked solution →Which of the following is the slope form of a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$?Read the answer and worked solution →For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, which of the following is the equation of the tangent at the point $(x_1,y_1)$ on the ellipse?Read the answer and worked solution →The area of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ isRead the answer and worked solution →If $\theta$ is the inclination of a line, its slope is defined as:Read the answer and worked solution →The slope of the x-axis is:Read the answer and worked solution →Two non-vertical lines are perpendicular if and only if the product of their slopes is:Read the answer and worked solution →

Worked practice questions

Equation of circle with center (h, k) and radius r:

  1. x + y = r
  2. x² + y² = h
  3. x² = r
  4. (x - h)² + (y - k)² = r²
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Correct answer: D. (x - h)² + (y - k)² = r²

Standard equation of circle. For center (0,0): x² + y² = r². **General second-degree:** x² + y² + 2gx + 2fy + c = 0, center (-g, -f), radius sqrt(g² + f² - c).

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Standard ellipse x²/a² + y²/b² = 1 (a > b) has eccentricity:

  1. e = sqrt(1 - b²/a²) (0 < e < 1)
  2. e > 1
  3. e = 1
  4. e = 0
Show answer and solution

Correct answer: A. e = sqrt(1 - b²/a²) (0 < e < 1)

e = sqrt(1 - b²/a²). For circle (a = b): e = 0. For ellipse: 0 < e < 1. Foci at (±ae, 0). As e → 1, ellipse becomes more elongated.

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Hyperbola x²/a² - y²/b² = 1 has eccentricity:

  1. e = sqrt(1 + b²/a²) > 1
  2. < 1
  3. 0
  4. Cannot determine
Show answer and solution

Correct answer: A. e = sqrt(1 + b²/a²) > 1

**For hyperbola:** e = sqrt(1 + b²/a²), always > 1. Foci at (±ae, 0). Asymptotes y = ±(b/a)x. **Rectangular hyperbola:** a = b, e = sqrt(2).

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