Daily questions _191207

Known elliptical \ (C: \ dfrac X ^ {2} + {A} ^ 2 \ dfrac ^ {Y} {2} = 2. 1 B ^ \) \ ((A> B> 0) \) about the focus are \ (of F_1, F_2 \) , elliptical \ (C \) centrifugation was \ (\ dfrac. 1} {2} {\) , and an elliptical \ (C \) through the point \ (\ left (1, - \ dfrac {. 3} {2} \ right) \) .
\ ((. 1) \) seeking elliptical \ (C \) is the standard equation;
\ ((2) \) when the straight line \ (L \) through elliptical \ ( C \) left vertex \ (M \) , and the ellipse \ (C \) is the intersection of the other \ (N \) , the straight line \ (NF_2 \) ellipse \ (C \) is the intersection of the other \ (P \) , if the \ (PF_1 \ PERP the MN \) , the Line \ (L \) equation.
Analysis:
\ ((1) \)Easy to get the questions on the \ (a, b \) equations \ [\ dfrac {1} { a ^ 2} + \ dfrac {9} {4b ^ 2} = 1, \ dfrac {b ^ 2} {a ^ 2} = \ dfrac {3
} {4}. \] Solutions have \ (\ left (A, B \ right) = \ left (2, \ sqrt {. 3} \ right) \) . whereby the elliptic equation seek is \ (\ dfrac {X ^ 2} {. 4} + \ dfrac {Y ^ 2} {. 3} =. 1 \) .
\ ((2) \) from the title, provided \ (\ angle NF_2x = \ theta \) , focal radius of the ellipse by the formula \ (\ rm {II} \ ) available
\ (| NF_2 | = \ {dfrac. 3} {2+ \ COS \ Theta} \) , whereby

Available \ (N \) coordinates of the point \ (N \ left (\ dfrac {3 \ cos \ theta} {2+ \ cos \ theta} +1, \ dfrac {3 \ sin \ theta} {2+ \ cos \ Theta} \ right) \) , Similarly available \ (P \) coordinates \ (P \ left (\ dfrac {-3 \ cos \ theta} {2- \ cos \ theta} +1, \ dfrac { -3 \ SiN \ Theta} {2- \ COS \ Theta} \ right) \) , because \ [PF_1 \ perp MN \ Rightarrow
\ overrightarrow {PF_1} \ cdot \ overrightarrow {MN} = 0. \] finishing available About \ (\ Theta \) equation \ [(7 \ cos \ theta
-5) \ left (\ cos \ theta + 1 \ right) = 0. \] Since \ (M, N \) points do not coincide, Thus \ (\ COS \ Theta +. 1 \ NEQ 0 \) , so that if and only if \ (\ cos \ theta = \ dfrac {5} {7} \) corresponding to the straight line \ (L \) is also desired. and the straight line equation \ [l: \ sqrt {6 } x-12y + 2 \ sqrt {6} = 0 \.]

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Origin www.cnblogs.com/Math521/p/11991763.html