Answer to Question #204747 in Calculus for Anuj

Question

Find the equation of the normal to the parabola y ² +4x = 0 at the point where the line y = x+c touches it.

Answer 1

Point of intersection of line on parabola,

$(x+c)^2+4x=0$

$x^2+c^2+2xc+4x=0$

$x^2+2x(c+2)+c^2=0$

$x=-(c+2)\pm\sqrt{2(2+c)}$

$y=-2+\sqrt{2(2+c)}$

Slope of Tangent to parabola

$2y\dfrac{dy}{dx}+4=0$

$m_T=\dfrac{dy}{dx}=\dfrac{-2}{y}=\dfrac{-2}{-2+\sqrt{2(2+c)}}$

Now,

Slope of normal = $m_N$

$m_T\times m_N=-1$

$m_N=\dfrac{-2+\sqrt{2(2+c)}}{2}$

Equation of normal,

$y+2-\sqrt{2(2+c)}=\dfrac{-2+\sqrt{2(2+c)}}{2}(x+(c+2)-\sqrt{2(2+c)})$