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

The correct option is A centreLet the two points be #160; A(a cosθ_1, b sinθ_1) and #160; B(a cosθ_2, b sinθ_2). O (0,0) is origin. #160;S ...

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Standard XII

Mathematics

Reflection of a Point about a Line

If tan θ1 t...

Question

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

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end of the minor axis

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Solution

The correct option is A centre
Let the two points be $A (a cos θ_{1}, b sin θ_{1})$ and $B (a cos θ_{2}, b sin θ_{2}) .$
O $(0, 0)$ is origin.
Slope of $O A = \frac{b sin θ_{1}}{a cos θ_{1}} = \frac{b tan θ_{1}}{a}$
Slope of $O B = \frac{b sin θ_{2}}{a cos θ_{2}} = \frac{b tan θ_{2}}{a}$
Slope of $O A \times$ slope of $O B = \frac{b^{2}}{a^{2}} tan θ_{1} tan θ_{2} = - 1$
Therefore, chord subtends right angle at origin.

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

The correct option is A centreLet the two points be A(acosθ1,bsinθ1) and B(acosθ2,bsinθ2).O (0,0) is origin. Slope of OA=bsinθ1acosθ1=btanθ1aSlope of OB=bsinθ2acosθ2=btanθ2aSlope of OA× slope of OB=b2a2tanθ1tanθ2=−1Therefore, chord subtends right angle at origin.

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