Conic Sections

For an ellipse with eccentricity e=1/2, focus at origin, and directrix x=4, the semi-major axis a satisfies a = e * distance from center to directrix. The center is at (-4/3, 0), so the directrix is at x = 4 (a/e to the right of center), giving a/e + a*e = 4, or a*(1/(1/2) + 1/2) = 4. Solving: (2a + a/2) = 4, so 5a/2 = 4, and a = 8/3. The ellipse equation x²/(64/9) + y²/(16) = 1 confirms this.

The eccentricity of a hyperbola and its conjugate are related by 1/e^2 + 1/e'^2 = 1. Applying the function f(e) repeatedly toggles between the two eccentricities. The integral of the constant values over the interval [1, 3] yields the result.

$\begin{array}{l} 3{ x^{ 2 } }+4{ y^{ 2 } }=1 \ Equation\, of\, \tan  gent\, to\, ellipse\, at\, \left( { { x _{ 1 } },{ y _{ 1 } } } \right)  \ 3x{ x _{ 1 } }+4y{ y _{ 1 } }=1 \ \frac { { 3x } }{ { \sqrt { 7 }  } } +\frac { { 4y } }{ { \sqrt { 7 }  } } =1 \ { x _{ 1 } }=\frac { 1 }{ { \sqrt { 7 }  } } \, \, \, \, \, { y _{ 1 } }=\frac { 1 }{ { \sqrt { 7 }  } }  \ \left( { \frac { 1 }{ { \sqrt { 7 }  } } ,\frac { 1 }{ { \sqrt { 7 }  } }  } \right)  \ Hence, \ option\, \, A\, is\, correct\, answer. \end{array}$

For an ellipse x^2/a^2 + y^2/b^2 = 1, the tangent with slope m is y = mx +/- sqrt(a^2m^2 + b^2). For a concentric circle x^2 + y^2 = r^2, the tangent is y = mx +/- r*sqrt(1+m^2). Equating these leads to the vertical tangent case.

The equation of the normal at $P\left( a\sec { \theta  } ,b\tan {

\theta  }  \right) $ to the hyperbola $\cfrac { { x }^{ 2 } }{ { a

}^{2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is
$ax\cos { \theta  } +by\cot { \theta  } ={ a }^{ 2 }+{ b }^{ 2 }$
This intersects the transverse and conjugate axes at $ L\left( \cfrac { {

a}^{ 2 }+{ b }^{ 2 } }{ a } \sec { \theta  } ,0 \right) $ and

$M\left(0,\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ { b }^{  } } \tan {

\theta  }  \right) $ respectively
Let $N(h,k)$. then $h=\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ b } \sec { \theta  } $ and
$k=\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ { b }^{  } } \tan { \theta  } $
$\Rightarrow

\sec { \theta  } =\cfrac { 2ah }{ { a }^{ 2 }+{ b }^{ 2 } } \quad

,\quad \tan { \theta  } =\cfrac { 2bk }{ { a }^{ 2 }+{ b }^{ 2 } }

\quad$
$\therefore \sec ^{ 2 }{ \theta  } -\tan ^{ 2 }{ \theta  }

=1\quad \Rightarrow 4{ a }^{ 2 }{ h }^{ 2 }-4{ b }^{ 2 }{ k }^{ 2 }={

\left( { a }^{ 2 }+{ b }^{ 2 } \right)  }^{ 2 }$
Thus the locus of

$(h,k)$ is $\Rightarrow 4{ a }^{ 2 }{ x }^{ 2 }-4{ b }^{ 2 }{ y }^{

2}={ \left( { a }^{ 2 }+{ b }^{ 2 } \right)  }^{ 2 }\quad $
Let ${ e } _{ 1 }$ be the eccentricity of this hyperbola. Then
${{

e } _{ 1 } }^{ 2 }=1+\cfrac { { a }^{ 2 } }{ { b }^{ 2 } } =\cfrac { {a

}^{ 2 }+{ b }^{ 2 } }{ { b }^{ 2 } } =\cfrac { { a }^{ 2 }{ e }^{ 2 }}{ {

a }^{ 2 }({ e }^{ 2 }-1) } $
$\Rightarrow { e } _{ 1 }=\displaystyle \frac { e }{ \sqrt { { e }^{ 2 }-1 }  } $

Suppose $\displaystyle \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ be a hyperbola and let $\displaystyle \dfrac{x^2}{a^2} - \frac{y^2}{b^2} = - 1$ be its conjugate.
Then their eccentricities are given by $e^2 = \displaystyle \dfrac{a^2 + b^2}{a^2}$ and $\displaystyle e'^2 = \frac{a^2 + b^2}{b^2}$ respectively.
$\therefore \displaystyle \dfrac{1}{e^2} + \dfrac{1}{e'^2} = \dfrac{a^2}{a^2 + b^2} + \dfrac{b^2}{a^2 + b^2} = 1$

The problem involves geometric properties of an ellipse. Solving for the orthocenter condition leads to the relation involving the eccentricity squared.

Given foci (3,1) and (1,1), the center is (2,1) and 2ae = 2, so ae = 1. Using the point (1,3), the sum of distances to the foci equals 2a.