Todai Entrance Exam: Math 2015 – Problem 2

(1)

Proof for foci of a ellipse

Proof for foci of a hyperbola is similar, with few notes:

Let h = \sqrt{c^{2} + d^{2}}:

    \[ \begin{Bmatrix} x \geq c: \frac{(xh + c^{2}) - (xh - c^{2})}{c} = 2c\\  x < -c: \frac{(- xh + c^{2}) - (- xh - c^{2})}{c} = 2c \end{matrix} \]

(2)

    \[ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{x}{a^{2}} + \frac{y}{b^{2}}\frac{\mathrm{d} y}{\mathrm{d} x} = 0 \Leftrightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = -\frac{x}{y}\frac{b^{2}}{a^{2}} \]

    \[ (***) \Leftrightarrow \frac{x^{3}}{y}\frac{b^{4}}{a^{4}} - \frac{x^{3}}{y}\frac{b^{2}}{a^{2}} + xy\frac{b^{2}}{a^{2}} + \frac{u^{2}x}{y}\frac{b^{2}}{a^{2}} - xy = 0 \]

    \[ \Leftrightarrow \frac{x^{3}}{y}\frac{b^{2}}{a^{2}}\left ( \frac{b^{2}}{a^{2}} - 1 \right ) + xy\left ( \frac{b^{2}}{a^{2}} - 1 \right ) + \frac{u^{2}x}{y}\frac{b^{2}}{a^{2}} = 0  \]

    \[ \Leftrightarrow \left ( \frac{b^{2}}{a^{2}} - 1 \right )\left ( \frac{x^{3}}{y}\frac{b^{2}}{a^{2}} + xy \right ) + \frac{u^{2}x}{y}\frac{b^{2}}{a^{2}} = 0 \]

    \[ \Leftrightarrow -\frac{u^{2}}{a^{2}}\left ( \frac{x^{3}b^{2} + xy^{2}a^{2}}{ya^{2}} \right )  + \frac{u^{2}x}{y}\frac{b^{2}}{a^{2}} = 0 \]

    \[ \Leftrightarrow -\frac{x^{3}b^{2} + xy^{2}a^{2}}{a^{2}} + xb^{2} = 0 \]

It’s worth noting that:

    \[ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Leftrightarrow y^{2}a^{2} = a^{2}b^{2} - b^{2}x^{2} \]

So:

    \[ -\frac{x^{3}b^{2} + xy^{2}a^{2}}{a^{2}} + xb^{2} = 0 \Leftrightarrow -\frac{x^{3}b^{2} + x(a^{2}b^{2} - b^{2}x^{2})}{a^{2}} + xb^{2} = 0 \Leftrightarrow - xb^{2} + xb^{2} = 0 \]

So the statement follows.

(3)

Similar to question (2)

(4)

Let (x_{0}, y_{0}) be a point on an ellipse in E_{u}. If a line is tangent to the ellipse at the point (x_{0}, y_{0}), then it has slope y_{e}'(x_{0}) = -\frac{x_{0}}{y_{0}}\frac{b^{2}}{a^{2}}, and the line perpendicular to the ellipse at that point has slope y_{d}'(x_{0}) = -\frac{1}{y_{e}'(x_{0})}. It is given from (***) that:

    \[ xyy'_{e}^{2} + (x^{2} - y^{2} - u^{2})y'_{e} - xy = 0 \Leftrightarrow xy + (x^{2} - y^{2} - u^{2})\frac{1}{y'_{e}} - xy\frac{1}{y'_{e}^{2}} = 0 \Leftrightarrow xy - (x^{2} - y^{2} - u^{2})y'_{d} - xyy'_{d}^{2} = 0 \]

(5)

I’m not sure if the equation in (4) is correct so I would rather not solve this.


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