InterviewSolution
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InterviewSolution
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P and Q are two points on the ellipse `(x^(2))/(a^(2)) +(y^(2))/(b^(2)) =1` whose eccentric angles are differ by `90^(@)`, thenA. Locus of point of intersection of tangents at P and Q is `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =2`B. Locus of mid-point `(P,Q)` is `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =(1)/(2)`C. Product of slopes of OP and OQ ehere O is the centre is `(-b^(2))/(a^(2))`D. Max. area of `DeltaOPQ` is `(1)/(2)ab` |
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Answer» Correct Answer - A::B::C::D `P =(a cos theta, b sin theta)` and `Q = (-a sin theta, b cos theta)` Tangent at `P, (x cos theta)/(a) + (y sin theta)/(b) =1` (1) `(-x sin theta)/(a) + (y cos theta)/(b) =1` (2) Elimination `theta`, by squaring and adding `(1)^(2) + (2)^(2) rArr (x^(2))/(a^(2)) + (y^(2))/(b^(2)) =2`, which is required locus. `:.` (a) is correct Now mid `(PQ) = ((a(cos theta-sin theta))/(2)(b(sin theta+cos theta))/(2)) =(x,y)` `cos theta - sin theta = (2x)/(a)` (3) `cos theta + sin theta = (2y)/(b)` (4) Squaring and adding (3) and (4), we get locus as `(x^(2))/(a^(2)) + (y^(2))/(b^(2)) =(1)/(2)` `:.` (b) is correct Slope of `OP = (b sin theta)/(a cos theta) = m_(1)` Slope of `OQ = (-b cos theta)/(a sin theta) = m_(2)` `:. m_(1)m_(2) = (-b^(2))/(a^(2))` `:.` (c) is correct Now are of triangle `OPQ = (1)/(2) |{:(a cos theta, b sin theta),(-a sin theta,b cos theta):}|=(1)/(2)ab(cos^(2)theta+sin^(2)theta)=(1)/(2)ab` `:.` (d) is corrext |
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