Pure Myths (2001 coor geo)

(a)P1(0,3) and P2(0,-5) are two points of C such that P1F and P2F are vertical
So two horizontal tangents are y=2 or y=-2

putting y=1 into the circle we have x=2√3 or -2√3
So the two vertical tangents are x=√3 or x=-√3

The equation of ellipse E is

(x/√3)^2+(y/2)^2=1
x^2/3+y^2/4=1
4x^2+3y^2=12

(b)
(i)
sub m=3p/(7-q), n=4(q-1)/(7-q) into 4x^2+3y^2
4x^2+3y^2
=4[3p/(7-q)]^2+3[4(q-1)/(7-q)]^2
=4[9p^2/(7-q)^2]+3[16(q-1)^2/(7-q)^2]
=[36p^2+48(q-1)^2]/(7-q)^2
=[36(16-(q+1)^2)+48(q-1)^2]/(7-q)^2
=[36(16-(q+1)^2)+48(q-1)^2]/(7-q)^2
=[576-36q^2-72q-36+48q^2-96q+48]/(7-q)^2
=[12q^2-168q+588]/(7-q)^2
=12(7-q)^2/(7-q)^2
=12

So M always lies on E

slope of FP is (q-1)/p

The equation of the tangent for the ellipse at M
4xm+3yn=12
sub m=3p/(7-q), n=4(q-1)/(7-q)
4x3p/(7-q)+3y4(q-1)/(7-q)=12
px+(q-1)y=(7-q)

slope is -p/(q-1)

The mid point of FP is (p/2,(q+1)/2)
sub into px+(q-1)y

p(p/2)+(q-1)(q+1)/2
=(p^2+q^2-1)/2
=(16-q^2-2q-1+q^2-1)/2
=7-q

So the tang at M to E is the prep bisector of FP

(c)
Now assume that we have a ellipse E 4x^2+3y^2=12
M(m,n) on ellipse with m=3p/(7-q), n=4(q-1)/(7-q)
we can solve to get
p=8m/(n+4) q=(7n+4)/(n+4)

Now
p^2+(q+1)^2
=(64m^2+64n^2+128n+64)/(n+4)^2
=(192-48n^2+64n^2+128n+64)/(n+4)^2
=(16n^2+128n+256)/(n+4)^2
=16

P(p,q) is on the C

the mid point of FP is 4m/(n+4),4(n+1)/(n+4)

slope of FP=3n/4m

The equation of the bisector of FP is
y-4(n+1)/(n+4)=(-4m/3n)(x-4m/(n+4))
(n+4)y-4(n+1)=(-4m/3n)((n+4)x-4m)
3n(n+4)y-12n(n+1)=(-4m)(n+4)x+16m^2
4m(n+4)x+3n(n+4)y=12n(n+1)+4[12-3n^2]
4m(n+4)x+3n(n+4)y=12n+48
4mx+3ny=12

which is the equation of the tangent to E at M

i think your question is self answered.

defined L is prep. bisector of FP.

a ellipse E was defined as all L tangent to it.

thus
any M on the ellipse. must lie on one of the L, which is a tangent line fo E and prep. bisector pf FP

the question is already been answered by it definition.

(p,q) is any pt P on C

Have you drawn a figure?
You either miss many details or misinterpret the question.
What is p and q ?