Let $A (α)$ and $B (β)$ be the extremities of a chord of an ellipse . If the slope of AB is equal to the slope of the tangent at a point $C (θ)$ on the ellipse, then the value of $θ$ , is:

\frac{α + β}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{α - β}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{α + β}{2} + π

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{α - β}{2} - π

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A $\frac{α + β}{2}$
C $\frac{α + β}{2} + π$
We have Slope of AB = Slope of tangent at C
$\Rightarrow \frac{b (sin β - sin α)}{a (cos β - cos α)} = \frac{- b cos θ}{a sin θ}$
$\Rightarrow \frac{- cos (\frac{α + β}{2})}{sin (\frac{α + β}{2})} = \frac{- cos θ}{sin θ}$
$\Rightarrow t a n (\frac{α + β}{2}) = t a n θ \Rightarrow θ = \frac{α + β}{2} + n π (n \in I)$
Hence, options 'A' and 'C' are correct.

Suggest Corrections

Similar questions

α

β

cos (\frac{α - β}{2}) =

(α, β)

y^{2} = 4 x

(α, β)

y^{2} = 4 x

y^{2} = x

(α, β), (β > 0)

x^{2} + 2 y^{2} = 1

α

L e t α, β b e s u c h t h a t π < α - β < 3 π . I f s i n α + s i n β = \frac{21}{65} a n d c o s α + c o s β = - \frac{27}{65}, t h e n t h e v a l u e o f c o s \frac{α - β}{2} i s

Join BYJU'S Learning Program