Show that the normals at the points of contact meet on the line $\frac{y}{y_1}=\frac{x}{x_1}$

Question

The tangents from $(x_1,y_1)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ intersect at right angles.Show that the normals at the points of contact meet on the line $\frac{y}{y_1}=\frac{x}{x_1}$

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The tangents from $(x_1,y_1)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ intersect at right angles on the director circle of ellipse $x^2+y^2=a^2+b^2$.The equation of chord of contact of tangents from point $(x_1,y_1)$ is $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$.
I am stuck here.

Answer 1

Equation of ellipse:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \tag{1}$$

Let $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$ and $D(x_4,y_4)$ be the points of the rectangle such that $A$ is on the director circle, $BD$ is the polar (i.e. the chord) of $A$ and $C$ is the required intersection.

As diagonals of a parallelogram bisect each other, $$(x_1,y_1)+(x_3,y_3)=(x_2,y_2)+(x_4,y_4)$$

Equation of chord (the polar):

$$\frac{x_1 x}{a^2}+\frac{y_1 y}{b^2}=1$$

$$y=\frac{b^2}{y_1}\left( 1-\frac{x_1 x}{a^2} \right) \tag{2}$$

Substitute $(2)$ into $(1)$,

$$\frac{x^2}{a^2}+\frac{b^2}{y_1^2} \left( 1-\frac{x_1 x}{a^2} \right)^2=1$$

$$\left( \frac{1}{a^2}+\frac{b^2 x_1^2}{a^4 y_1^2} \right)x^2- \frac{2b^2 x_1}{a^2 y_1^2}x+\frac{b^2}{y_1^2}-1=0$$

Sum of roots:

\begin{align} x_2+x_4 &= \frac{2a^2b^2 x_1}{a^2 y_1^2+b^2 x_1^2} \\ x_3 &= \frac{2a^2b^2 x_1}{a^2 y_1^2+b^2 x_1^2}-x_1 \\ &= \frac{2a^2b^2-a^2 y_1^2-b^2 x_1^2}{a^2 y_1^2+b^2 x_1^2}x_1 \\ \end{align}

Smilarly,

$$y_3=\frac{2a^2b^2-a^2 y_1^2-b^2 x_1^2}{a^2 y_1^2+b^2 x_1^2}y_1$$

Hence,

$$\frac{y_3}{x_3}=\frac{y_1}{x_1}$$

Answer 2

If the tangents are normal to each other, the normals are normal to them, then the four lines make a quadrilateral with three, thus four, right angles, i.e. a rectangle.

You already computed the equation of the diagonal line, so you know the tangent points $P_2$ and $P_3$.

Then compute the line parallel to $P_1,P_2$ through $P_3$, and the other parallel to $P_1,P_3$ and passing though $P_2$. Cross them to get $P_4$.