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Angle of Intersection of Two Curves

The point of ...

Question

The point of intersection of tangents to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ at two points

the sum of whose eccentric angles is constant, is $(a \frac{c o s \frac{α + β}{2}}{c o s \frac{α - β}{2}}, b \frac{s i n \frac{α + β}{2}}{c o s \frac{α - β}{2}})$

A

True

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B

False

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Solution

The correct option is A True
Point of intersection is $(a \frac{c o s \frac{α + β}{2}}{c o s \frac{α - β}{2}}, b \frac{s i n \frac{α + β}{2}}{c o s \frac{α - β}{2}})$

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Similar questions

α - β =

constant, then the locus of the point of intersection of tangents at

P (a cos α, b sin α)

Q (a cos β, b cos β)

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

Q. If the tangents to the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

α

β

with the major axis such that

tan α + tan β = λ

, then the locus of their point of intersection is

Q. The locus of point of intersection of tangents drawn at

α, β

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

α - β = \frac{π}{3}

α

β

are the eccentric angles of points of contract of tangents drawn from (3,2) to the ellipse

\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1

| t a n (\frac{α - β}{2}) | =

Q. If the chord through the points whose eccentric angles are

α

β

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

passes through the focus

(a e, 0),

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\frac{β}{2} =

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