Assertion (A): The difference of the slopes o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The difference of the slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{2 \sqrt{h^2 - ab}}{|b|}$.For the given equati

Vedclass · Accepted Answer

Both $A$ and $R$ are true and $R$ is the correct explanation of $A$

Vedclass · Answer

(A) The difference of the slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{2 \sqrt{h^2 - ab}}{|b|}$.For the given equation $y^2 - 2xy \sec^2 \alpha + (3 + 	an^2 \alpha)(	an^2 \alpha - 1) x^2 = 0$,we have $a = (3 + 	an^2 \alpha)(	an^2 \alpha - 1)$,$2h = -2 \sec^2 \alpha$ (so $h = -\sec^2 \alpha$),and $b = 1$.The difference of slopes is $\frac{2 \sqrt{(-\sec^2 \alpha)^2 - (3 + 	an^2 \alpha)(	an^2 \alpha - 1)(1)}}{|1|}$.$= 2 \sqrt{\sec^4 \alpha - (3	an^2 \alpha - 3 + 	an^4 \alpha - 	an^2 \alpha)}$.$= 2 \sqrt{\sec^4 \alpha - (	an^4 \alpha + 2	an^2 \alpha - 3)}$.Since $\sec^2 \alpha = 1 + 	an^2 \alpha$,then $\sec^4 \alpha = (1 + 	an^2 \alpha)^2 = 1 + 2	an^2 \alpha + 	an^4 \alpha$.Substituting this,the expression becomes $2 \sqrt{1 + 2	an^2 \alpha + 	an^4 \alpha - 	an^4 \alpha - 2	an^2 \alpha + 3} = 2 \sqrt{4} = 4$.Thus,both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

Assertion $(A)$: The difference of the slopes of the lines represented by $y^2 - 2xy \sec^2 \alpha + (3 + \tan^2 \alpha)(\tan^2 \alpha - 1) x^2 = 0$ is $4$.
Reason $(R)$: The difference of the slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is $\frac{2 \sqrt{h^2 - ab}}{|b|}$.

Similar Questions

The equation of the pair of lines passing through the origin and parallel to the lines $y = m_1x + c_1$ and $y = m_2x + c_2$ is:

The lines represented by the equation $a{x^2}(b - c) - xy(a(b - c) + c(a - b)) + c{y^2}(a - b) = 0$ are

If the equation $hxy + gx + fy + c = 0$ represents a pair of straight lines,then

If the lines represented by the equation $2x^2 - pxy + 2y^2 = 0$ are real,then the value of '$p$' lies in the interval

If the slope of one line of $ax^2+4xy+y^2=0$ is $3$ times the other,then the value of '$a$' is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion $(A)$: The difference of the slopes of the lines represented by $y^2 - 2xy \sec^2 \alpha + (3 + \tan^2 \alpha)(\tan^2 \alpha - 1) x^2 = 0$ is $4$.Reason $(R)$: The difference of the slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is $\frac{2 \sqrt{h^2 - ab}}{|b|}$.