Amit M Agarwal Solutions for Chapter: Ellipse, Exercise 4: Target Exercises

If $CP$ and $CD$ are conjugate semi-diameters of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then, the locus of the midpoint of $PD$ is

The locus of poles of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with respect to concentric ellipse $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1,$ is

The locus of pole of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with respect to the parabola $y^2=4ax$, is

The locus of the poles of normal chords of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ is

If $CP$ and $CD$ are conjugate semi-diameters of the ellipse, then $PD$ touches the ellipse

If $\alpha$ and $\beta$ are the angles subtended by the major axis of an ellipse at the extremities of a pair of conjugate diameters, then

If $PC{P}^{\text{'}}$ and $DC{D}^{\text{'}}$ form a pair of conjugate diameters of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $R$ is any point on the circle $x^2+y^2=c^2,$ then

$PC{P}^{\text{'}}$ and $DC{D}^{\text{'}}$ are conjugate diameters of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\theta$ is the eccentric angle of $P.$ Then eccentric angle of the point where the circle $P{P}^{\text{'}}D$ again cuts the ellipse