Single Choice

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 $$(-3\sin \theta, \sqrt {3}\cos \theta); \theta \in \left (0, \dfrac {\pi}{2}\right )$$; then $$\dfrac {2\cot \beta}{\sin 2\theta}$$ is equal to

A$$\sqrt {2}$$
B$$\dfrac {2}{\sqrt {3}}$$
Correct Answer
C$$\dfrac {1}{\sqrt {3}}$$
D$$\dfrac {\sqrt {3}}{4}$$

Solution

x2+3y2=9

⇒2x+6y
dy
dx

=0 ....... Differentiating w.r.t x

⇒
dy
dx

=
−x
3y


Equation of normal is

−
dx
dy

=
3y
x



dx
dy

|(3cosθ,
√
3
sinθ)=
3
√
3
sinθ
−3cosθ

=
√
3
tanθ=m1

dx
dy

|(−3sinθ,
√
3
cosθ)=
3
√
3
cosθ
−3sinθ

=−
√
3
cotθ=m2

β is the angle between the normals to the ellipse (i), then

tanβ=|
m1−m2
1+m1m2

|

=|
√
3
tanθ+
√
3
cotθ
1−3tanθcotθ

|

=|
√
3
tanθ+
√
3
cotθ
1−3

|

tanβ=
√
3
2

|tanθ+cotθ|

1
cotβ

=
√
3
2

|tanθ+cotθ|

1
cotβ

=
√
3
2

|
sinθ
cosθ

+
cosθ
sinθ

|

1
cotβ

=
√
3
2

|
1
sinθcosθ

|

1
cotβ

=
√
3
sin2θ



⇒
2cotβ
sin2θ

=
2
√
3

SIMILAR QUESTIONS

Ellipse

If line $$y = 2x + c$$ is a normal to the ellipse $$\dfrac {x^{2}}{9} + \dfrac {y^{2}}{16} = 1$$, then find the value of c.

Ellipse

A point on the ellispe $$x^{2}+ 3y^{2}=37$$ where the normal is parallel to the line 6x-5y=2 is

Ellipse

If from a point P$$(0, \alpha)$$ two normals other than axes are drawn to ellipse $$\dfrac{x^{2}}{25}+ \dfrac{y^{2}}{16}=1 $$, such that $$|\alpha|$$

Ellipse

Show that the locus of the intersection of two perpendicular normals to an ellipse is the curve $$\left( { a }^{ 2 }+{ b }^{ 2 } \right) \left( { x }^{ 2 }+{ y }^{ 2 } \right) { \left( { a }^{ 2 }{ y }^{ 2 }+{ b }^{ 2 }{ x }^{ 2 } \right) }^{ 2 }={ \left( { a }^{ 2 }-{ b }^{ 2 } \right) }^{ 2 }{ \left( { a }^{ 2 }{ y }^{ 2 }-{ b }^{ 2 }{ x }^{ 2 } \right) }^{ 2 }$$.

Ellipse

ABC is a triangle inscribed in the ellipse $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ having each side parallel to the tangent at the opposite angular point; prove that the normals at A, B, and C meet at a point which lies on the ellipse. $$a^2x^2+b^2y^2=\dfrac{1}{4}(a^2-b^2)^2$$.

Ellipse

If the normal at $$P$$ meet the axis in $$G$$ and $$g$$, then $$PG=Pg=pc$$.

Ellipse

Prove that the straight line $$lx + my = n$$ is a normal to the ellipse, if $$\dfrac {a^{2}}{l^{2}} + \dfrac {b^{2}}{m^{2}} = \dfrac {(a^{2} - b^{2})^{2}}{n^{2}}$$.

Ellipse

If $$CP$$ be conjugate to the normal at $$Q$$, prove that $$CQ$$ is conjugate to the normal at $$P$$.

Ellipse

If the normal to the ellipse $$3x^2+4y^2=12$$ at a point P on it is parallel to the line, $$2x+y=4$$ and the tangent to the ellipse at P passes through $$Q(4, 4)$$ then PQ is equal to?

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