Parametric Form of Normal : Ellipse

Q.

Consider a hyperbola $H:x^2–2y^2=4$. Let the tangent at a point $P(4,\sqrt{6})$ meet the x-axis at $Q$ and latus rectum at $R(x_1,y_1),x_1>0$. If $F$ is a focus of $H$ which is nearer to the point $P$, then the area of $\Delta QFR$ is equal to:

1. $\sqrt{6}-1$

2. $4\sqrt{6}-1$

3. $4\sqrt{6}$

4. $\frac{7}{\sqrt{6}}-2$

Q. If $Ax+By=5$ is a normal at a point $P$ on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ whose eccentric angle is $\frac{\pi }{4}$, then the value of $(A+B{)}^2$ is

Q. Any ordinate $MP$ of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ meets the auxiliary circle at $Q$, then locus of the point of intersection of normals at $P$ and $Q$ to the respective curves is

1. $x^2+y^2=8$
2. $x^2+y^2=34$
3. $x^2+y^2=64$
4. $x^2+y^2=16$

Q. If the tangent drawn at point $P(t^2,2t)$ on the parabola $y^2=4x$ is same as the normal drawn at point $Q(\sqrt{5}\cos \theta ,2\sin \theta )$ on the ellipse $4x^2+5y^2=20,$ then

1. $Q\equiv (-1,\frac{4}{\sqrt{5}})$
2. $Q\equiv (-1,-\frac{4}{\sqrt{5}})$
3. $P\equiv (\frac{1}{5},-\frac{2}{\sqrt{5}})$
4. $P\equiv (\frac{1}{5},\frac{2}{\sqrt{5}})$

Q. If the normal at the point $P\left(\theta \right)$ to the ellipse $\frac{x^2}{14}+\frac{y^2}{5}=1$ intersects it again at the point $Q\left(2\theta \right)$, then

1. $\sin \theta =-\frac{2}{3}$
2. $\sin \theta =\frac{3}{4}$
3. $\cos \theta =\frac{3}{4}$
4. $\cos \theta =-\frac{2}{3}$

Q. The normal at a point $P$ on the ellipse $x^2+4y^2=16$ meets the $x-$ axis at $Q$. If $M$ is the midpoint of the line segment $PQ$, then the locus of $M$ intersects the latus rectum of the given ellipse at the points

1. $(\pm \frac{3\sqrt{5}}{2},\pm \frac{2}{7})$
2. $(\pm \frac{3\sqrt{5}}{2},\pm \frac{\sqrt{19}}{7})$
3. $(\pm 2\sqrt{3},\pm \frac{1}{7})$
4. $(\pm 2\sqrt{3},\pm \frac{4\sqrt{3}}{7})$

Q. If the normal at $\theta$ on the ellipse $5x^2+14y^2=70$ cuts the curve again at a point $2\theta$, then $\cos \theta$ =

1. $\frac{2}{3}$
2. $-\frac{2}{3}$
3. $\frac{1}{3}$
4. $-\frac{1}{3}$

Q. If the normal at any point $P$ on the ellipse $\frac{x^2}{9}+\frac{y^2}{16}=1$ meets the axes at $G$ and $g,$ respectively, then the ratio of $PG:Pg$ is

1. $3:4$
2. $4:3$
3. $9:16$
4. $16:9$

Q. If normal at a variable point $P$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ of eccentricity $e$ meets the axes of the ellipse at $Q$ and $R,$ then the locus of the midpoint of $QR$ is a conic with an eccentricity $e^{\prime }$ such that

1. $e^{\prime }$ is independent of $e$
2. $e^{\prime }=2e$
3. $e^{\prime }=e$
4. $e^{\prime }=\frac{e}{2}$