Question

Write down the equations to the tangent and normal at the point of the parabola $y^2=6x$ whose ordinate is 12,

Answer 1

$y^2=6xy=12{\left(12\right)}^2=6x\Rightarrow x=24P(24,12)$

Equation of tangent at $(x_1,y_1)$ to $y^2=4ax$ is

$y{y}_1=4a\left(\frac{x+x_1}{2}\right)$

So, the equation of tangent at $P$ to $y^2=6x$ is

$y\left(12\right)=6\left(\frac{x+24}{2}\right)4y=x+244y-x=24$

Slope of tangent $m=-\frac{a}{b}=-\frac{-1}{4}=\frac{1}{4}$

Let the slope of normal be $m^{\prime }$

$m{m}^{\prime }=-1\frac{1}{4}{m}^{\prime }=-1\Rightarrow m^{\prime }=-4$

Equation of normal at $P$ is

$y-12=-4(x-24)y+4x=108$