If tan-1-tan-2-a2-b2 then the chord joining two points -1 and -2 on the ellipse x2-a2-y2-b2-1 will subtend a right angle at-

The correct option is C Centre Let P be (θ_1) and Q =(θ_2) Slope OP =b/a tan θ_1 and slope OQ =b/a tan θ_2 ∴M_OP×M_OQ = b^2/a^2× ( a^2/b^2 )= 1 ∴ It subtends ...

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Byju's Answer

Standard XII

Physics

Law of Conservation of Mechanical Energy

If tanθ1·tanθ...

Question

If $tan θ_{1} . tan θ_{2} = - \frac{a^{2}}{b^{2}}$ then the chord joining two points $θ_{1}$ and $θ_{2}$ on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ will subtend a right angle at:

Focus

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Centre

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End of the major axis

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End of the minor axis

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Solution

The correct option is B
Centre

Let P be $(θ_{1})$ and Q $= (θ_{2})$

Slope OP $= \frac{b}{a} tan θ_{1}$

and slope OQ $= \frac{b}{a} tan θ_{2}$

$∴ M_{OP} \times M_{OQ} = \frac{b^{2}}{a^{2}} \times (\frac{- a^{2}}{b^{2}}) = - 1$

$∴$ It subtends $90^{\circ}$ at centre.

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Similar questions

Q. If

tan θ_{1} tan θ_{2} = - \frac{a^{2}}{b^{2}}

, then the chord joining two points

θ_{1}

and

θ_{2}

on the ellipse

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

will subtend a right angle at

Q. If tan

θ_{1}

tan

θ_{2} = - \frac{a^{2}}{b^{2}},

then the chord joining two points

θ_{1}

θ_{2}

on the ellipse

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

will subtend a right angle at

Q. If the eccentric angles of the ends of a focal chord of the ellipse

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

are

θ_{1}

and

θ_{2}

, then value of

tan \frac{θ_{1}}{2} \times tan \frac{θ_{2}}{2}

equals to:

Q. Two tangents to the ellipse

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

make angles

θ_{1}, θ_{2}

with the major axis. If

tan θ_{1}, tan θ_{2}

is a constant

^{'} k^{'}

then the equation to the locus of their point of intersection is

Q. If the chord joining the points

α

and

β

on the ellipse

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

subtends a right angle at (a, 0), then

tan (\frac{α}{2}) tan (\frac{β}{2}) =

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If tan θ1. tan θ2=−a2b2 then the chord joining two points θ1 and θ2 on the ellipse x2a2+y2b2=1 will subtend a right angle at:

The correct option is B Centre Let P be (θ1) and Q =(θ2) Slope OP =ba tan θ1 and slope OQ =ba tan θ2 ∴MOP×MOQ=b2a2×(−a2b2)=−1 ∴ It subtends 90∘ at centre.

If $tan θ_{1} . tan θ_{2} = - \frac{a^{2}}{b^{2}}$ then the chord joining two points $θ_{1}$ and $θ_{2}$ on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ will subtend a right angle at:

The correct option is B
Centre

Let P be $(θ_{1})$ and Q $= (θ_{2})$

Slope OP $= \frac{b}{a} tan θ_{1}$

and slope OQ $= \frac{b}{a} tan θ_{2}$

$∴ M_{OP} \times M_{OQ} = \frac{b^{2}}{a^{2}} \times (\frac{- a^{2}}{b^{2}}) = - 1$

$∴$ It subtends $90^{\circ}$ at centre.