Slope Form of Normal : Ellipse

Q. A point on the ellipse $x^2+3y^2=37$ where the normal is parallel to the line $6x-5y=2$ is

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

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.

What is the equation of the normal which is perpendicular to 3x + 4y = 5 for the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

1. $y=\frac{4}{3}x-\frac{4(a^2+b^2)}{\sqrt{9a^2+16b^2}}$

2. $y=\frac{4}{3}x-\frac{a^2-b^2}{\sqrt{16a^2+9b^2}}$

3. $y=\frac{4}{3}x-\frac{4(a^2+b^2)}{\sqrt{16a^2+9b^2}}$

4. $y=\frac{4}{3}x-\frac{4(a^2-b^2)}{\sqrt{9a^2+16b^2}}$

Q. If $\beta$ is one of the angles between the normals to the ellipse, $x^2+3y^2=9$ at the points $(3\cos \theta ,\sqrt{3}\sin \theta )$ and $(-3\sin \theta ,\sqrt{3}\cos \theta )$; $\theta \in (0,\frac{\pi }{2});$ then $\frac{2\cot \beta }{\sin 2\theta }$ is equal to :

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

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. $\frac{(a^2+b^2{)}^2}{2(a^2+b^2)}$