Would anyone mind telling me how to solve (a)? I have no idea what I should do to solve this problem.
Also, what is principal axes?
[Math] Ellipse, hyperbola and principle axis
analytic geometryconic sections
Related Solutions
Focal semi axis is equal to the length of the semi axis passing through focus of ellipse or hyperbole. Focal semi axis mean same as semi major axis. I hope you can now get answers as: For ellipse: $a^2=49$ and $b^2=36$. For hyperbola: $a^2=9$ and $b^2=4$ Hope it helps.
With the given parametric expressions we have the derivative of the coordinates with respect to the parameter as $$\left\{ \begin{aligned} x &= a \cos\theta \\ y &= b \sin\theta \end{aligned} \right. \quad \implies \left\{ \begin{aligned} \dot{x} &= -a \sin\theta \\ \dot{y} &= b \cos\theta \end{aligned} \right. \qquad \text{where}~~ \dot{x} \equiv \frac{dx}{d\theta},~\dot{y} \equiv \frac{dy}{d\theta}~.$$ Thus the slope of the tangent line at point $P = (a\cos\theta, b\sin\theta)$ is $$\frac{dy}{dx} = \frac{ \dot{y} }{ \dot{x} } = -\frac{ b\cos\theta}{a\sin\theta}~,$$ and the slope of the normal line is the negative inverse of that: $$m \equiv -\left( \frac{dy}{dx} \right)^{-1} = \frac{ -\dot{x} }{ \dot{y} } = \frac{ a\sin\theta}{ b\cos\theta}~.$$ Note that here the parameter happens to be the polar angle, therefore $m$ is just the $\tan\theta$ (the slope of a normal line to a circle at $\theta$ angle) scaled by $\frac{a}b$.
The equation to the normal line is thus $$ \frac{ y - b\sin\theta }{ x - a\cos\theta } = m = \frac{ a\sin\theta}{ b\cos\theta}~,$$ or equivalently $$y - b \sin\theta = \frac{ a\sin\theta}{ b\cos\theta} (x - a\cos\theta)~ \tag*{Eq.(1)} $$ or to compare it with what is stated in the question post: $$\frac{ y \cos\theta }a - \frac{ x \sin\theta }b = \frac{ b^2 - a^2}{ ab} \sin\theta \cos\theta ~.$$
Denote $(x_0, y_0)$ as the coordinates of point $N$, the intersection of the normal line with the major axis (which is also the $x$-axis). The $x$-coordinate is readily calculated with $\text{Eq.}\,\text{(1)}$ at $y_0 = 0$
\begin{align} && y_0 - b \sin\theta &= \frac{ a\sin\theta}{ b\cos\theta} (x_0 - a\cos\theta) \\ &\implies & \frac{ - b^2 }a \cos\theta &= x_0 - a\cos\theta \\ &\implies & x_0 &= \left( a - \frac{ b^2 }a \right) \cos\theta \end{align} Denoting $r \equiv a - \frac{ b^2 }a = \frac{ a^2 - b^2 }a $, we have the center of the line segment $\overline{PN}$ being at $$ \left( \frac{ x + x_0}2 ~, \frac{ y + y_0}2\right) = \left( \frac{ a + r}2 \cos\theta ~, \frac{ b}2 \sin\theta\right) = \left( \frac{ 2a^2 - b^2}{ 2 a } \cos\theta ~, \frac{ b}2 \sin\theta\right) $$ At this point, we can identify the requested locus as an ellipse centere at the same origin and parametrized (necessarily) by the same parameter $\theta$ with the principle axes $$a' \equiv \frac{ 2a^2 - b^2}{ 2 a} = a - \frac{b^2}{ 2a } ~~, ~~b' \equiv \frac{b}2$$
Further Discussion
Note that the ratio $\frac{ a' }{ b'} \neq \frac{a}b$. That is, the new ellipse is NOT similar to the original. Nor does the ratio equal to $\sqrt{ \frac{a}b }$ or $\frac{a^2}{ b^2}$ or anything particularly nice. What we can say about the shape is that this new ellipse (from the locus) is more "elongated" than the original. Consider the wide-and-short original ellipse where $a > b$, we have $$ \frac{ a' }{ b'} = \frac{2a}b - \frac{b}a = \frac{a}b + \left( \frac{a}b - \frac{b}a \right) > \frac{a}b $$ where the larger ratio gives a more stretched out ellipse.
Also note that the locus goes "outside" of the original two foci, $$a' > \sqrt{a^2 - b^2} \qquad \because (a')^2 = a^2 - b^2 + \frac{ b^4 }{ 4a^2 } > a^2 - b^2$$ with its own foci that are closer to the center. $$(a')^2 - (b')^2 = a^2 - b^2 + \frac{ b^2}4 \left( \frac{ b^2 }{ a^2 } - 1\right) < a^2 - b^2$$
Best Answer
I think $x_1$ and $x_2$ are the co-ordinate axis . I am going by the following approach , which is a little long but this method works .
Seeing the $x_1x_2$ term I figured out that the principal axis of this curve are not perpendicular to the co-ordinate axes . Hence I supposed the equation of the curve of the form $$a(x+ty)^2+b(tx-y)^2=1$$
Now I have taken that the principal axes pass through the origin because there are no $x_1$ and $x_2$ term of single power . And the principal axes should be perpendicular to each other hence I have taken the slope of one to be $t$ and other to be $-\frac{1}{t}$ (the product of slopes of two perpendicular lines is equal to $-1$)
Now we are all set to go , expanding the above equation we will get ( work out the steps , I am writing the final result )
$$(a+bt^2)x_{1}^2+2t(a-b)x_1x_2+(at^2+b)x_{2}^2 = 1$$
Now reducing the original equation to this form you will get $$ \frac{6}{14}x_1^2-\frac{4}{14}x_1x_2+\frac{3}{14}x_2^2 =1 $$
Equating the co-efficients you will get $$a+bt^2 = \frac{6}{14}\ \ \ \ \ \ \ \ \ \ \ \ \ \cdots(1)$$ $$2t(a-b)=-\frac{4}{14}\ \ \ \ \ \ \cdots(2)$$ $$at^2+b=\frac{3}{14}\ \ \ \ \ \ \ \ \ \ \ \ \ \cdots(3)$$
Now subtract $(3)$ from $(1)$ you will get an equation in $(a-b)$ and $t$ . Using the $(2)$ equation put the value $(a-b)$ in terms of $t$ you will get an equation in $t$ , then use that value of $t$ in $(1)$ and $(2)$ to get $a$ and $b$ .
This way you get the value of $t$ hence the equation of principal axes and value of $a$ and $b$ come to be positive , hence it is an ellipse .