Home
Class 12
MATHS
If the centre of the hyperbola whose equ...

If the centre of the hyperbola whose equation is `ax^(2) + 2hxy + by^(2) + 2gx + 2fy + c = 0 " be " (alpha, beta)` , then find the equation of the asymptotes.

Text Solution

AI Generated Solution

The correct Answer is:
To find the equation of the asymptotes of the hyperbola given by the equation \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) with center at \( (\alpha, \beta) \), we can follow these steps: ### Step 1: Understand the equation of the hyperbola The general form of the hyperbola is given by: \[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \] This equation can represent a hyperbola that is rotated and translated from the origin. ### Step 2: Identify the center of the hyperbola The center of the hyperbola is given as \( (\alpha, \beta) \). This means that we need to find the asymptotes of the hyperbola at this center. ### Step 3: Formulate the function \( f(x, y) \) Define the function: \[ f(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c \] The asymptotes can be derived from the condition that the function behaves like a combination of lines. ### Step 4: Set up the equation for asymptotes The asymptotes can be found using the expression: \[ f(x, y) - f(\alpha, \beta) = 0 \] This means we need to evaluate \( f(\alpha, \beta) \): \[ f(\alpha, \beta) = a\alpha^2 + 2h\alpha\beta + b\beta^2 + 2g\alpha + 2f\beta + c \] ### Step 5: Substitute into the asymptote equation Now, substituting this back into our asymptote equation gives: \[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c - (a\alpha^2 + 2h\alpha\beta + b\beta^2 + 2g\alpha + 2f\beta + c) = 0 \] The constant \( c \) cancels out: \[ ax^2 + 2hxy + by^2 + 2gx + 2fy - (a\alpha^2 + 2h\alpha\beta + b\beta^2 + 2g\alpha + 2f\beta) = 0 \] ### Step 6: Simplify the equation Let \( k = a\alpha^2 + 2h\alpha\beta + b\beta^2 + 2g\alpha + 2f\beta \). Thus, we can rewrite the equation of the asymptotes as: \[ ax^2 + 2hxy + by^2 + 2gx + 2fy - k = 0 \] ### Step 7: Identify the asymptotes The asymptotes of the hyperbola can be obtained by setting the quadratic terms to zero: \[ ax^2 + 2hxy + by^2 = 0 \] This represents the equations of the asymptotes. ### Final Equation of Asymptotes To find the explicit equations of the asymptotes, we can factor the quadratic equation: \[ ax^2 + 2hxy + by^2 = 0 \] This can be solved using the quadratic formula or by finding the roots, leading to the equations of the asymptotes.
Promotional Banner

Topper's Solved these Questions

  • HYPERBOLA

    FIITJEE|Exercise Exercise - 3|6 Videos
  • HYPERBOLA

    FIITJEE|Exercise ASSIGNMENT PROBLEMS ( SUBJECTIVE) Level - I|15 Videos
  • HYPERBOLA

    FIITJEE|Exercise Exercise - 1|6 Videos
  • HEIGHTS & DISTANCE

    FIITJEE|Exercise ASSIGNMENT PROBLEMS (OBJECTIVE) LEVEL-II|20 Videos
  • INDEFINTE INTEGRAL

    FIITJEE|Exercise EXERCISE-8|1 Videos

Similar Questions

Explore conceptually related problems

the equation ax^(2)+ 2hxy + by^(2) + 2gx + 2 fy + c=0 represents an ellipse , if

The principal axes of the hyperbola ax^(2) + 2hxy + by^(2) + 2gx + 2fy + c = 0 are parallel to the lines

The equation ax^(2)+2hxy+by^(2)+2gx+2fy+c=0 represents a circle if

The order of the differential equation whose solution is x^(2)+y^(2)+2gx+2fy+c=0 is

The equation ax^(2)+2hxy+by^(2)+2gx+2fy+c=0 repersents a hyperbola if

For the equation ax^(2) +by^(2) + 2hxy + 2gx + 2fy + c =0 where a ne 0 , to represent a circle, the condition will be

If the equation ax^(2)+2hxy+by^(2)+2gx+2fy+c=0 represents a pair of parallel lines, then

Find the centre and radius of the circle ax^(2) + ay^(2) + 2gx + 2fy + c = 0 where a ne 0 .