Question

The value of parameter ‘a’ so that the line $(3 - a)x + ay + {a^2} - 1 = 0$ is a normal to the curve $xy = 1$ is/are:
A) $(3,\infty )$
B) $( - \infty ,0)$
C) $(0,3)$
D) None of these

Answer 1

For solving this particular question, we first consider the slope of the normal and then we have to find the slope of the given line then compare it and lastly we have to consider the cases to solve this particular question.

Complete step by step solution:
Now, we know that slope of the normal at point say $P({x_1},{y_1})$ is ${x^2}$ ,
And here we also know that the slope of the normal is always greater than or equal to zero.
We have given a line that is $(3 - a)x + ay + {a^2} - 1 = 0$ , This type of linear equations sometimes called as slope-intercept form because we can easily find the slope ,
Slope of the given line is $\dfrac{{a - 3}}{a} \geqslant 0$
Here we have two cases:
Case I: when both the numerator and denominator are negative or
$a - 3 \leqslant 0$ and $a < 0$
$\Rightarrow a \in [3,\infty )$
Case II: when both the numerator and denominator are positive or
$a - 3 \geqslant 0$ and $a > 0$
$\Rightarrow a \in ( - \infty ,0)$
From both the cases we have $a \in ( - \infty ,0)\bigcup {[3,\infty )}$ .
Hence, option ‘A’ and ‘B’ are correct. Both the options are correct.

Note:
Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily. This type of linear equation is sometimes called slope-intercept form because we can easily find the slope and the intercept of the corresponding lines. This also allows us to graph it. We can quickly tell the slope i.e., $m$ the y-intercepts i.e., $(y,0)$ and the x-intercept i.e., $(0,y)$ .we can graph the corresponding line.