Slope Form of Normal : Ellipse

Q. Number of distinct normals, that can be drawn to the ellipse $\frac{x^2}{169}+\frac{y^2}{25}=1$ from the point $P(0,6)$ is

1. one
2. two
3. three
4. four

Q. The tangent and normal to the ellipse $3x^2+5y^2=32$ at the point $P(2,2)$ meet the $x-$axis at $Q$ and $R$, respectively. Then the area (in sq. units) of the triangle $PQR$ is :

1. $\frac{34}{15}$
2. $\frac{16}{3}$
3. $\frac{14}{3}$
4. $\frac{68}{15}$

Q. A normal inclined at an angle of $\frac{\pi }{4}$ to the x-axis of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is drawn. It meets the major and minor axes in P and Q respectively. If C is the centre of the ellipse then the area of the triangle CPQ is

1. $\frac{(a^2-b^2{)}^2}{4(a^2+b^2)}$
2. $\frac{(a^2-b^2{)}^2}{(a^2-b^2)}$
3. $\frac{(a^2-b^2{)}^2}{2(a^2+b^2)}$
4. $\frac{(a^2+b^2{)}^2}{2(a^2+b^2)}$

Q. Coordinates of point(s) on the ellipse $x^2+3y^2=37,$ where the normal is parallel to the line $6x-5y=2,$ is/are

1. $(5,-2)$
2. $(5,2)$
3. $(-5,2)$
4. $(-5,-2)$