# Conic curve having 5 points concurrent normals at the origin.

Consider the curve represented by the equation $ax^2+2bxy+cy^2+d=0$ in the plane where $a>0 ,c>0$ and $ac>b^2$.Suppose that the normals to the curve drawn at $5$-distinct points on the curve all pass through the origin.Then,

$(a)$ $a=c$ and $b>0$.

$(b)$ $a \ne c$ and $b=0$.

$(c)$ $a \ne c$ and $b<0$.

$(d)$ None of the above.

My attempt $:$

Let $(x_{1}, y_{1})$ be one of the points on the given curve at which the normal to the given curve passes through the origin.Now the equation of the normal at the point $(x_{1} , y_{1})$ is :

$(y-y_1)(ax_1 +by_1) =(bx_1 + cy_1)(x-x_1)$.

As it passes through the origin so we have $b({x_{1}}^2 - {y_{1}}^2)-(a-c)x_1 y_1 = 0$.

This shows that the points on the given curve at which normals drawn pass through the origin will lie on the curve $b({x}^2 - {y}^2)-(a-c)x y = 0$.

So the given curve and the curve just obtained has five common points.Since the curves are quadratic in $x$ and $y$ having five common points.So these two curves should coincide (because $5$ constraints will determine $4$-unknowns uniquely).Now equating the coefficients of like terms of these two equations we have :

$ab+bc=0 , 2b^2 = a(c-a)$ and $bd=0$.Since $a>0,c>0$ we must have $a+c>0$ and hence $b=0$.Hence from the second equation we have $c=a$ since $a>0$.

So the correct option should be $a=c$ and $b=0$.Hence $(a),(b)$ and $(c)$ are all incorrect.Hence $(d)$ is the only correct option.

Is the above reasoning correct at all?Please verify it.

• If we put $x=0$ and $y=0$ in the equation of the normal passing through $(x_{1} , y_{1})$ then what do we get?How can we get $\pm$?Would you please explain me? – Arnab Chatterjee. May 31 '17 at 7:12