Diameter of Ellipse - Practice Questions & MCQ

Diameter of 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.

Let $(\mathrm{h}, \mathrm{k})$ be the mid-point of the chord $\mathrm{y}=\mathrm{m} \mathrm{x}+\mathrm{c}$ 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 PQ and RS be 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}
$