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P and Q are two points on the ellipse (x...

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^(@)`, then

A

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`

Text Solution

Verified by Experts

The correct Answer is:
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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