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If the straight line 4ax+3by=24 is a no...

If the straight line `4ax+3by=24` is a normal to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 (agtb)`, then find the the coordinates of focii and the ellipse

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The correct Answer is:
`(4+-sqrt(10),0)`
Given normal is `4ax+3by=34 " "(1)`
Let this line be normal to the ellipse at `P(a cos theta, b sin theta)` is `ax sec theta-"by cosec" theta= a^(2)-b^(2)" "(2)`
Equations (1) and (2) are identical, so `(sectheta)/(4)=(-"cosec"theta)/(3)=(a^(2)-b^(2))/(24)`
`:. cos theta=(6)/(a^(2)-b^(2))and sin theta=(-8)/(a^(2)-b^(2))`
Squaring and adding, we get
`(64)/((a^(2)-b^(2))^(2))+(36)/((a^(2)-b^(2))^(2))=1`
`rArr 100=(a^(2)-b^(2))^(2)`
`rArra^(2)-b^(2)=10" "( :. agtb)`
`rArra^(2)e^(2)=10`
Hence, foci are `(+-sqrt(10),0)`
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