Podcast
Questions and Answers
Prove that 4p ^ 2 + q ^ 2 = a ^ 2 for the given equation x * \cos \alpha - y * \sin \alpha = a * \cos 2\alpha
Prove that 4p ^ 2 + q ^ 2 = a ^ 2 for the given equation x * \cos \alpha - y * \sin \alpha = a * \cos 2\alpha
The length p is the distance from the origin to the line, and q is the distance from the origin to the line. Using the given equation, we can derive the values of p and q, and then substitute them into the equation 4p ^ 2 + q ^ 2 = a ^ 2 to prove the relationship.
For the equation a * x ^ 2 + 2hxy + b * y ^ 2 + 2gx + 2fy + c = 0 representing two parallel lines, prove that (i) h ^ 2 = ab (ii) a * f ^ 2 (iii) the distance between the parallel lines is 2\sqrt{\frac{g ^ 2 - ac},{a(a + b)}} or 2\sqrt{\frac{r ^ 2 - bc},{b(a + b)}}
For the equation a * x ^ 2 + 2hxy + b * y ^ 2 + 2gx + 2fy + c = 0 representing two parallel lines, prove that (i) h ^ 2 = ab (ii) a * f ^ 2 (iii) the distance between the parallel lines is 2\sqrt{\frac{g ^ 2 - ac},{a(a + b)}} or 2\sqrt{\frac{r ^ 2 - bc},{b(a + b)}}
(i) To prove h ^ 2 = ab, we can use the condition for the lines to be parallel and express h in terms of a, b, and h. (ii) To prove a * f ^ 2, we can use the equation and express f in terms of a, b, g, and c. (iii) The distance between the parallel lines can be found using the formula 2\sqrt{\frac{g ^ 2 - ac},{a(a + b)}} or 2\sqrt{\frac{r ^ 2 - bc},{b(a + b)}}.
Show that the lines joining the origin to the points of intersection of the curve x ^ 2 - xy + y ^ 2 + 3x + 3y - 2 = 0 and the straight line x - y - \sqrt{2} = 0 are mutually perpendicular.
Show that the lines joining the origin to the points of intersection of the curve x ^ 2 - xy + y ^ 2 + 3x + 3y - 2 = 0 and the straight line x - y - \sqrt{2} = 0 are mutually perpendicular.
We can find the points of intersection of the curve and the straight line, then calculate the slopes of the lines joining the origin to these points. If the product of the slopes is -1, we can conclude that the lines are mutually perpendicular.
Find the angle between two non-parallel lines whose direction cosines satisfy the equations 6mn - 2nl + 5 / m = 0.
Find the angle between two non-parallel lines whose direction cosines satisfy the equations 6mn - 2nl + 5 / m = 0.
Signup and view all the answers
If y = \tan(4x-4x^2+1-6x^2+x), show that \rac{d},{dx} (y) = \rac{1},{1 + x ^ 2}
If y = \tan(4x-4x^2+1-6x^2+x), show that \rac{d},{dx} (y) = \rac{1},{1 + x ^ 2}
Signup and view all the answers
Study Notes
Proofs and Derivations
- The equation x * cos(α) - y * sin(α) = a * cos(2α) can be proved to be equal to 4p^2 + q^2 = a^2.
Properties of Parallel Lines
- For the equation ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0, representing two parallel lines:
- h^2 = ab
- a*f^2
- Distance between the parallel lines is 2√((g^2 - ac) / (a(a + b))) or 2√((r^2 - bc) / (b(a + b))).
Mutual Perpendicularity of Lines
- The lines joining the origin to the points of intersection of the curve x^2 - xy + y^2 + 3x + 3y - 2 = 0 and the straight line x - y - √2 = 0 are mutually perpendicular.
Angle between Non-Parallel Lines
- The angle between two non-parallel lines whose direction cosines satisfy the equations 6mn - 2nl + 5 / m = 0 can be found.
Trigonometric Derivatives
- For y = tan(4x - 4x^2 + 1 - 6x^2 + x), the derivative dy/dx = 1 / (1 + x^2).
Studying That Suits You
Use AI to generate personalized quizzes and flashcards to suit your learning preferences.
Description
Test your knowledge of geometry and algebraic equations with this challenging quiz. Prove the relationships between line equations, lengths of perpendiculars, and distances between parallel lines.
