Question

If the line `l x+m y+n=0` cuts the ellipse `((x^2)/(a^2))+((y^2)/(b^2))=1` at points whose eccentric angles differ by `pi/2,` then find the value of `(a^2l^2+b^2m^2)/(n^2)` .

Answer 1

Let the line `lx+ my+n=0` cut the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 "at" P(a cos theta, b sin theta) and Q (a cos ((pi)/(2)+theta),b sin((pi)/(2)+theta))= Q(-a sin theta, b cos theta)`
Since points P and Q lie on the line, we have
`la cos theta+mb sin theta=-n`
and `-la cos theta+mb sin theta=-n`
Squaring and adding , we get
`a^(2)l^(2)+b^(2)m^(2)=2n^(2)`
`rArr(a^(2)l^(2)+b^(2)m^(2))/(n^(2))=2`