Question

Three normals to y^2=4x pass through the points (15,12)

Answer 1

I am solving this question for you

Answer 2

$4x-y-72=0$ this is the equation.

Step-by-step explanation:

Given,

Three normal to $y^{2}=4x$ pass through the points $(15,12)$

We know,

Normal equation is,

$y=mx-2am-am^{3}$__1

Compare $y^{2}=4x$ this equation with parabola equation $y^{2}=4ax$

Then,

 $4a=4$

⇒$a=1$

For points $(15,12)$,

$12=15m-2m-m^{3}$

⇒$m^{3} -13m-12=0$

By calculating above equation we can find the value of $m$ is,

$m=-1,m=-3 and m=4$

By neglecting negative value,

$m=4$

Plug the value in equation-1,

$y=4x-2(1)(4)-(1)(4)^{3}$

⇒$y=4x-8-64$

∴$4x-y-72=0$

So, $4x-y-72=0$ this is the equation.