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Question: If \({\mathrm{tan\mathrm\theta}}_1{\mathrm{tan\mathrm\theta}}_2=-\dfrac{\mathrm a^2}{\mathrm b^2}\),...

If tanθ1tanθ2=−a2b2{\mathrm{tan\mathrm\theta}}_1{\mathrm{tan\mathrm\theta}}_2=-\dfrac{\mathrm a^2}{\mathrm b^2}, then the chord joining two points θ1  and  θ2{\mathrm\theta}_1\;\mathrm{and}\;{\mathrm\theta}_2 on the ellipse x2a2+y2b2=1\dfrac{\mathrm x^2}{\mathrm a^2}+\dfrac{\mathrm y^2}{\mathrm b^2}=1 will subtend a right angle at-
a)Focus
b)Centre
c)End of major axis
d)End of minor axis

Explanation

Solution

Hint: To solve this problem, one should have knowledge about the parametric form of an ellipse. The parametric form of a general ellipse at point θ\mathrm\theta is given by-
(acosθ,  bsinθ)\left(\mathrm{acos\mathrm\theta},\;\mathrm{bsin\mathrm\theta}\right)
The focus is the point about which a conic section is constructed. An ellipse has two foci. The centre is the geometrical centre inside the ellipse. The end of the major axis is the end point of the longer diameter of the ellipse. The end of the minor axis is the end point of the shorter diameter.

Complete step-by-step answer:

This is the ellipse with equation x2a2+y2b2=1\dfrac{\mathrm x^2}{\mathrm a^2}+\dfrac{\mathrm y^2}{\mathrm b^2}=1. Let the point E and F be at an angle θ1  and  θ2{\mathrm\theta}_1\;\mathrm{and}\;{\mathrm\theta}_2 respectively.
By parametric form of ellipse, the coordinates of point E and F are-
E(acosθ1,  bsinθ1)F(acosθ2,  bsinθ2)\mathrm E\left({\mathrm{acos\mathrm\theta}}_1,\;{\mathrm{bsin\mathrm\theta}}_1\right)\\\\\mathrm F\left({\mathrm{acos\mathrm\theta}}_2,\;{\mathrm{bsin\mathrm\theta}}_2\right)
The slope of a point(p,q) with respect to origin is given by-
m=q−0p−0=qp\mathrm m=\dfrac{\mathrm q-0}{\mathrm p-0}=\dfrac{\mathrm q}{\mathrm p}
So, the slopes of points E and F with respect to origin are-
mE=bsinθ1acosθ1=batanθ1mF=bsinθ2acosθ2=batanθ2{\mathrm m}_{\mathrm E}=\dfrac{{\mathrm{bsin\mathrm\theta}}_1}{{\mathrm{acos\mathrm\theta}}_1}=\dfrac{\mathrm b}{\mathrm a}{\mathrm{tan\mathrm\theta}}_1\\\\{\mathrm m}_{\mathrm F}=\dfrac{{\mathrm{bsin\mathrm\theta}}_2}{{\mathrm{acos\mathrm\theta}}_2}=\dfrac{\mathrm b}{\mathrm a}{\mathrm{tan\mathrm\theta}}_2
If we multiply mEm_E and mFm_F, the result should be-
mE×mF=batanθ1×batanθ2b2a2ˉtanθ1tanθ2{\mathrm m}_{\mathrm E}\times{\mathrm m}_{\mathrm F}=\dfrac{\mathrm b}{\mathrm a}{\mathrm{tan\mathrm\theta}}_1\times\dfrac{\mathrm b}{\mathrm a}{\mathrm{tan\mathrm\theta}}_2\\\=\dfrac{\mathrm b^2}{\mathrm a^2}{\mathrm{tan\mathrm\theta}}_1{\mathrm{tan\mathrm\theta}}_2
It is given in the question that tanθ1tanθ2=−a2b2{\mathrm{tan\mathrm\theta}}_1{\mathrm{tan\mathrm\theta}}_2=-\dfrac{\mathrm a^2}{\mathrm b^2}, on substituting the value-
mE×mF=b2a2×−a2b2=−1m_E\times m_F=\dfrac{b^2}{a^2}\times-\dfrac{a^2}{b^2}=-1
If the product of slopes of two lines is -1, then the lines are perpendicular to each other. Hence, the two points E and F subtend a right angle at the centre of the ellipse.
The correct option is B. Centre

Note: To solve this problem, one should have knowledge about an ellipse as well as the slope of a line. The parametric form of a conic is very useful in solving problems as it reduces the number of variables and the calculation. If it is required to prove two lines to be perpendicular it is sufficient to prove the product of their slopes to be -1.