Diameter of Ellipse

What is the Diameter of the Ellipse?

The locus of the mid-points of a system of parallel chords of an ellipse is called a diameter and the point where the diameter intersects the ellipse is called the vertex of the diameter.
Locus of Mid Point

Let $(\mathrm{h}, \mathrm{k})$ be the mid-point of the chord $\mathrm{y}=\mathrm{m} x+\mathrm{c}$ of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then
$\mathrm{T}=\mathrm{S}_1 \quad$ [equation of chord bisected at given point]

$
\begin{aligned}
\Rightarrow & \frac{\mathrm{xh}}{\mathrm{a}^2}+\frac{\mathrm{yk}}{\mathrm{b}^2} =\frac{\mathrm{h}^2}{\mathrm{a}^2}+\frac{\mathrm{k}^2}{\mathrm{~b}^2} \\
\Rightarrow & \mathrm{k} =-\frac{\mathrm{b}^2 \mathrm{~h}}{\mathrm{a}^2 \mathrm{~m}}
\end{aligned}
$
Hence, the locus of the mid-point is $y=-\frac{b^2 x}{a^2 m}$

Conjugate Diameters

Two diameters are said to be conjugate when each bisects all chords parallel to the other.

If $y=m_1 x$ and $y=m_2 x$ be two conjugate diameters of an ellipse, then

$
m_1 m_2=-\frac{b^2}{a^2}
$
If $P Q$ and $R S$ are two conjugate diameters. Then the coordinates of the four extremities of two conjugate diameters are

$
\begin{aligned}
& P \equiv(a \cos \phi, b \sin \phi) \\
& Q \equiv(-a \cos \phi,-b \sin \phi) \\
& S \equiv(-a \sin \phi, b \cos \phi) \\
& R \equiv(a \sin \phi,-b \cos \phi)
\end{aligned}
$

Properties of diameters

1) The tangent at the extremity of any diameter is parallel to the chords it bisects or parallel to the conjugate diameter.
2) The tangents at the ends of any chord meet on the diameter which bisects the chord.

Properties of conjugate diameters
1) The eccentric angles of the ends of a pair of conjugate diameters of an ellipse differ by a right angle.
2) The sum of the square of any two conjugate semi-diameters of an ellipse is constant and equal to the sum of squares of the semi-axis.
3) The product of the focal distances of a point on an ellipse is equal to the square of the semi-diameter which is conjugate to the diameter through the point.
4) Two conjugate diameters are called equi conjugate if their lengths are equal.

Solved Examples Based on Diameter of Ellipse

Example 1: The Locus of midpoints of chords of the ellipse $\frac{x^2}{2}+y^2=1$ which are tangents to the ellipse $x^2+\frac{y^2}{2}=1$ is
Solution:
Locus of the mid-point of the chord of the ellipse $\frac{x^2}{2}+y^2=1_{\text {is }} y=-\frac{x}{2 m}$
The equation of a tangent to the ellipse $x^2+\frac{y^2}{2}=1$ in slope form is $y=m x \pm \sqrt{m^2+2}$
from eq (i) and eq (ii)

$
\begin{aligned}
& \Rightarrow y=\left(-\frac{x}{2 y}\right) x+\sqrt{\left(-\frac{x}{2 y}\right)^2+2} \\
& \Rightarrow \quad\left(2 y^2+x^2\right)^2=x^2+8 y^2
\end{aligned}
$
Hence, the correct answer is $\left(2 y^2+x^2\right)^2=x^2+8 y^2$

Example 2: If the product of focal distances of a point $\mathrm{P}(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)$ on an ellipse $\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ is $\lambda$-times the square of the semi-diameter CD (of conjugate diameter CP ), then $\lambda=$ Solution: Let foci be $\mathrm{S}(\mathrm{ae}, 0)$ and $\mathrm{S}^{\prime}(-\mathrm{ae}, 0) \mathrm{As} \mathrm{P}$ is $(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)$
D is $\left(\mathrm{a} \cos \left(\theta+\frac{\pi}{2}\right), \mathrm{b} \sin \left(\theta+\frac{\pi}{2}\right)\right)$

$
=(-\mathrm{a} \sin \theta, \mathrm{b} \cos \theta)
$
By using the definition of an ellipse,

$
\mathrm{PS}=\mathrm{e}(\mathrm{PM})
$
Or $\operatorname{PS}=\mathrm{e}\left(\frac{\mathrm{a}}{\mathrm{e}}-\mathrm{a} \cos \theta\right)$

$
\begin{aligned}
& =\mathrm{a}-\mathrm{ae} \cos \theta \\
& =\mathrm{a}(1-\mathrm{e} \cos \theta)(\text { Standard Result }) \\
& \therefore \quad \text { SP.S'P }=\mathrm{a}(1-\mathrm{e} \cos \theta) \mathrm{a}(1+\mathrm{e} \cos \theta) \\
& =\mathrm{a}^2\left(1-\mathrm{e}^2 \cos ^2 \theta\right)=\mathrm{a}^2-\mathrm{a}^2 \mathrm{e}^2 \cos ^2 \theta
\end{aligned}
$
But $b^2=a^2\left(1-e^2\right)$

$
\begin{aligned}
& \therefore \quad \text { SP.S'P }=\mathrm{a}^2-\left(\mathrm{a}^2-\mathrm{b}^2\right) \cos ^2 \theta \\
& =\mathrm{a}^2 \sin ^2 \theta+\mathrm{b}^2 \cos ^2 \theta=\mathrm{CD}^2
\end{aligned}
$

Hence, the correct answer is 1

Example 3: If one end of the diameter of the ellipse $4 x^2+y^2=16$ is $(\sqrt{3}, 2)$, then the other end is
Solution: Since every diameter of an ellipse passes through the centre and is bisected by it, therefore the coordinates of the other end are $(-\sqrt{3},-2)$
Hence, the answer is $(-\sqrt{3},-2)$

Example 4: A ray emanating from the point $(-0,3)$ is incident on the ellipse $16 \mathrm{x}^2+25 y^2=400$ at the point $P$ with ordinate 4 . Then the equation of the reflected ray after the first reflection is.
Solution: For point P y-coordinate $=4$
Given ellipse is $16 \mathrm{x}^2+25 y^2=400$

$
16 \mathrm{x}^2+25(4)^2=400, \therefore \mathrm{x}=0
$

co-ordinate of co-ordinate of P is $(0,4)$

$
\begin{aligned}
& e^2=1-\frac{16}{25}=\frac{9}{25} \\
& e=\frac{3}{5}
\end{aligned}
$

$\therefore$ Foci $( \pm \mathrm{ae}, 0)$, i.e. $( \pm 3,0)$
Equation of reflected ray
(i.e.PS) is $\frac{x}{3}+\frac{y}{4}=1$ or $4 x+3 y=12$.

Hence, the correct answer is $4 x+3 y=12$

Example 5: If the points of intersection of the ellipses $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{p^2}+\frac{y^2}{q^2}=1$ be the extremities of the conjugate diameters of first ellipse, then $\frac{a^2}{P^2}+\frac{b^2}{q^2}=$
Solution: Subtracting in order to find their points of intersection, we get

$
\mathrm{x}^2\left(\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{p}^2}\right)+\mathrm{y}^2\left(\frac{1}{\mathrm{~b}^2}-\frac{1}{\mathrm{q}^2}\right)=0
$
The above equation will represent a pair of conjugate diameters of the first ellipse if

$
\mathrm{m}_1 \mathrm{~m}_2=-\frac{\mathrm{b}^2}{\mathrm{a}^2}
$
But $\mathrm{m}_1 \mathrm{~m}_2=\frac{\mathrm{A}}{\mathrm{B}}=-\frac{\mathrm{b}^2}{\mathrm{a}^2}$

$
\begin{aligned}
& \therefore\left(\frac{1}{a^2}-\frac{1}{p^2}\right) \div\left(\frac{1}{b^2}-\frac{1}{q^2}\right)=-\frac{b^2}{a^2} \\
& \text { or } a^2\left(\frac{1}{a^2}-\frac{1}{p^2}\right)+b^2\left(\frac{1}{b^2}-\frac{1}{q^2}\right)=0 \\
& \text { or } \frac{a^2}{p^2}+\frac{b^2}{q^2}=2
\end{aligned}
$
Hence, the correct answer is 2