Answer to Question #200471 in Analytic Geometry for tanya

Question #200471

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


1
Expert's answer
2021-06-16T15:06:47-0400

Answer:-

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)})"

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