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If the normals at alpha, beta,gamma and ...

If the normals at `alpha, beta,gamma` and `delta` on an ellipse are concurrent then the value of `(sigma cos alpha)(sigma sec alpha)` I

A

2

B

4

C

6

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
Let the equation of the ellipse be
`(x^(2))/(a^(2)) + (y^(2))/(b^(2)) =1`
Equation of a normal at any point `P(theta)` on the ellipse is
`(a sec theta) x -(b cosec theta)y = a^(2)e^(2)`
Let normal is passing through a point `A(h,k)`
`rArr (ah sec theta - a^(2)e^(2))^(2) = (bk cosec theta)^(2)`
`rArr a^(2)h^(2) sec^(2) theta -2a^(3)e^(2)h sec theta +a^(4) e^(4)`
`= b^(2)k^(2) cosec^(2) theta = b^(2)k^(2) ((sec^(2)theta)/(sec^(2)theta-1))`
`rArr a^(2)h^(2) sec^(4) theta - 2a^(3) e^(2)h sec^(3) theta + (a^(4) e^(4) -a^(2)h^(2)-b^(2)k^(2))`
`sec^(2) theta + 2a^(3)a^(2)h sec theta - a^(4) e^(4) =0` (1)
If `alpha, beta, gamma` and `delta` be the roots of the above equation, then
`Sigma sec alpha =(2a^(3)e^(2)h)/(a^(2)h^(2)) = (2ae^(2))/(h)`
Multiplying equation (1) by `cos^(4) theta`, it reduces to
`a^(4)e^(4)cos^(4) theta - 2a^(3)e^(2)h cos^(3) theta - (a^(4)e^(4)-a^(2)h^(2)-b^(2)k^(2))`
`cos^(2) theta + 2a^(3) e^(2) h cos theta - a^(2)h^(2) = 0` (2)
Then `Sigma cos alpha = (2a^(3)e^(2)h)/(a^(4)e^(4)) = (2h)/(ae^(2))`
Hence, we have
`(Sigma sec alpha) (Sigma cos alpha) = (2ae^(2))/(h).(2h)/(ae^(2)) =4`
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