Find the number of normals that can be drawn through the point $(0,\frac12)$ to the ellipse $\frac{x^2}2+y^2=1$
My Attempt:
Normal equation is $\frac{\sqrt2x}{\cos\theta}-\frac y{\sin\theta}=1$, if $\theta$ is the eccentric angle of a point on the ellipse.
It passes through $(0,\frac12)$. So, $-\frac1{2\sin\theta}=1\implies\sin\theta=-\frac12\implies$ the point lies in 3rd or 4th quadrant.
So, there are $3$ normals. (One is y-axis)
Is this correct?
Here's a more algebraic approach, avoiding the trigonometric parameterization of the ellipse. With some modifications, this method applies to a conic over any field of characteristic $\ne 2$ in which square roots exist.First we observe that the only vertical normal to the curve is the $Y$ axis, which passes through the point(0,1/2) and the only horizontal normal to the curve is the $X$ axis, which does not pass through the point (0,1/2). Let us look for other normals. By implicit differentiation, we see that the slope of the tangent line to the ellipse at the point $(x_1,y_1)$ on the ellipse is $\frac{x_1}{-2y_1}$ , so the slope of the normal line at that point is $\frac{2y_1}{x_1}$. Note that we have already looked at the horizontal and vertical cases, so we can assume $x_1,y_1 \ne 0.$ The equation of the normal line is $$y-y_1=\frac{2y_1}{x_1}(x-x_1)$$. This line must pass through $(0,\frac{1}{2})$, so $$\frac{1}{2}-y_1=\frac{2y_1}{x_1}(-x_1)$$ $$=-2y_1.$$ Hence, $y_1=-\frac{1}{2}$. Since $(x_1,y_1)$ lies on the conic $$\frac{x^2}{2}+y^2=1,$$ we have $$\frac{x_1^2}{2}+\frac{1}{4}=1$$. Thus, $\frac{x_1^2}{2}=\frac{3}{4}, x_1= \pm \sqrt{\frac{3}{2}}.$ Thus a non-vertical normal to the conic through $(0,\frac{1}{2})$ passes through $(\sqrt{\frac{3}{2}},-\frac{1}{2})$ or $(-\sqrt{\frac{3}{2}},-\frac{1}{2})$, i.e. there are 3 normals to the conic.