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Assertion (A): The difference of the slopes of the lines represented by
$\mathrm{y}^2-2 \mathrm{xysec}^2 \alpha+\left(3+\tan ^2 \alpha\right)\left(-1+\tan ^2 \alpha\right) \mathrm{x}^2=0 \text { is } 4$
Reason (R): The difference of the slopes represented by
$a x^2+2 h x y+b y^2=0 \text { is } \frac{2 \sqrt{h^2-a b}}{|b|}$
Options:
$\mathrm{y}^2-2 \mathrm{xysec}^2 \alpha+\left(3+\tan ^2 \alpha\right)\left(-1+\tan ^2 \alpha\right) \mathrm{x}^2=0 \text { is } 4$
Reason (R): The difference of the slopes represented by
$a x^2+2 h x y+b y^2=0 \text { is } \frac{2 \sqrt{h^2-a b}}{|b|}$
Solution:
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Verified Answer
The correct answer is:
Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
Since the difference of the slopes of the lines $a x^2+2 h x y+b y^2=0$ is given by $\frac{2 \sqrt{h^2-a b}}{|b|}$ Now the given line is
$y^2-2 x y \sec ^2 \alpha+\left(3+\tan ^2 \alpha\right)\left(-1+\tan ^2 \alpha\right) x^2=0$
So the difference of slope is
$=\frac{2 \sqrt{\left(\sec ^2 \alpha\right)^2+\left(3+\tan ^2 \alpha\right)\left(\tan ^2 \alpha-1\right)}}{1}$
$=2 \sqrt{\sec ^2 \alpha\left(\sec ^2 \alpha+3-\tan ^2 \alpha\right)}=2 \sqrt{4}=4$
So $\mathrm{a}$ is correct option.
$y^2-2 x y \sec ^2 \alpha+\left(3+\tan ^2 \alpha\right)\left(-1+\tan ^2 \alpha\right) x^2=0$
So the difference of slope is
$=\frac{2 \sqrt{\left(\sec ^2 \alpha\right)^2+\left(3+\tan ^2 \alpha\right)\left(\tan ^2 \alpha-1\right)}}{1}$
$=2 \sqrt{\sec ^2 \alpha\left(\sec ^2 \alpha+3-\tan ^2 \alpha\right)}=2 \sqrt{4}=4$
So $\mathrm{a}$ is correct option.
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