Question

The equations $x=t^3+9$ and $y=\frac{3t^3}{4}+6$ represents a straight line where $t$ is a parameter. The $y$ intercept of the line is

• A. $-\frac{3}{4}$
• B. $9$
• C. $6$
• D. $1$

Answer 1

The correct option is A $-\frac{3}{4}$

$x=t^3+9$
$\Rightarrow t^3=x-9$
$y=\frac{3t^3}{4}+6$
$\Rightarrow y=\frac{3}{4}(x-9)+6$
For y-intercept , put $x=0$
$\Rightarrow y=-\frac{3}{4}$