Question

The equation(s) of tangents drawn from the point $(1,4)$ to the parabola $y^2=12x$ is/are

• A. $x-y+3=0$
• B. $x-y+1=0$
• C. $x-2y+4=0$
• D. $3x-y+1=0$

Answer 1

Answer: (A), (D)

$3x-y+1=0$

Parabola equation : $y^2=12x$
$\Rightarrow a=3$
Equation of tangent in slope form is
$y=mx+\frac{a}{m}$
$\Rightarrow y=mx+\frac{3}{m}$
This equation passes through $(1,4)$
$\Rightarrow 4=m+\frac{3}{m}$
$\Rightarrow 4m=m^2+3$
$\Rightarrow m^2-4m+3=0$
$\Rightarrow m^2-3m-m+3=0$
$\Rightarrow m(m-3)-1(m-3)=0$
$\Rightarrow (m-1)(m-3)=0$
So, $m=1,3$
Equation of tangent $:y=x+\frac{3}{1}$
$\Rightarrow x-y+3=0$
For $m=3$, $y=3x+\frac{3}{3}$
$\Rightarrow 3x-y+1=0$