Question

find the equation of the tangents through the point (8,-1) to the ellipse 2x^2+5y^2=70

Answer 1

1) Check that it is possible. Is (8,-1) actually OUTSIDE the ellipse?

Answer 2

Yes, (8,-1) is not on the ellipse, because \(5y^2\) is odd at y=-1,
and \(2x^2\) is even at any integer x (certain at an even x=8).

Even + Odd \(\ne\) Even

Answer 3

my graphing calculator gives this

Answer 4

So you have two tangents

Answer 5

As in, OUTSIDE, not just "not on".

Answer 6

Given some other point, other than (8,-1), say (a,b), what is the equation of the line containing both (8,1) and (a,b)?

Answer 7

the slope would be \[\frac{ b+1 }{ a-8 }\]so the equation would be y+1=(b+1/a-8)*(x-8)

Answer 8

like that?

Answer 9

@aliLnn
Let's first put the equation of the ellipse in order:
2x^2+5y^2=70
=>
\(\large \frac{x^2}{35}+\frac{y^2}{14}=1\)...................(1)
=> \(a^2=35, b^2=14\)...............(2)

Then, you can assume the line to have slope m, so the line passing through (8,-1)=(x0,y0) is
y-y0=m(x-x0),
or
y=mx+(y0-mx0)
\(y= mx+q\)..............................(3)
where \(q=(y0-mx0)\)..............(4)

At the point of tangency, it's a common point between the line and the ellipse, so we can substitute (3) into (1) and solve for x to get
\((a^2m^2+b^2)x^2+2a^2mqx+a^2q^2-a^2b^2=0\)............(5)
Let
\(A=a^2m^2+b^2\) ..........(6a)
\(B=2a^2mq\), and..............(6b)
\(C=a^2q^2-a^2b^2\)...............(6c)
(5) simplifies to
\(Ax^2+Bx+C=0\)...............................................................(5a)

The condition for tangency is such that there is a double root (multiplicity 2), or
B^2-4AC=0 ......................................(7)

Substitute (6a), (6b) and (6c) into (7) results in
\(-4a^2b^2(q^2-a^2m^2-b^2)=0\) ...........................(8)

Since \(a^2\ne 0\) and \(b^2\ne 0\), we have
\(q^2-a^2m^2-b^2=0\) ...........................................(8a)
from which we can solve for m (a^2, b^2, x0, y0 are all numeric quantities).
Can you solve the quadratic (8a) for m? There are two distinct roots.

Answer 10

By the way, q=q(m), so substitute equation (4) in (8a) before solving for m.
For checking: the solutions for m satisfy |m|\(\le\)1, or you can post your answers for checking.
After that, you would use equation (3) to find the tangent lines.