Let a tangent drawn at point P(other than vertex) on ellipse. If a line AP intersect the line passing through B perpendicular to above tangent at Q, then AQ is equal to (where A(-2,-2) and B(2,2)).
I converted given equation to general form $(x+y)^2/18 + (x-y)^2/2 =1$. After this I let P as (h,k) and then solved it further but unable to get the required answer Please give me solution to this problem
On
There is a simple geometric reason due to the fact that $A$ and $B$ are the foci of the ellipse which can be described in an alternate way under the classical geometric definition of the locus:
$$\{ P \ | \ PA+PB=6 \}\tag{1}$$
(see sketch of proof below).
In this way, point $Q$ being symmetrical of point $B$ with respect to the tangent, $BPQ$ is an isosceles triangle ; as a consequence $PQ=PB$. Therefore, using (1):
$$AQ=AP+PQ=AP+PB=6$$
Sketch of proof of (1): It suffices to show that the given equation is equivalent to (1) under the form:
$$PA+PB=\sqrt{(x+2)^2+(y+2)^2}+\sqrt{(x-2)^2+(y-2)^2}=6$$
This is done by squaring, giving rise to an expression where there is a single remaining square root ; isolating it, say on the LHS, and squaring again, one gets the intial equation of the ellipse.
Another way would be to compute the lengths $a=3, b=1$ of the semiaxes, then obtain the focal distance using formula $f=\sqrt{a^2-b^2}=2\sqrt{2}$ which is indeed the value of $OA=OB$.
We know that the standard form of an ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
We can observe the lines $x=0$ and $y=0$ from the equation of the ellipse, which are nothing but the axes of the ellipse.
WLOG, we can conclude that whenever we write an ellipse as $\frac{ax+by+c}{A^2}+\frac{px+qy+r}{B^2}=1$ where $ax+by+c=0$ and $px+qy+r=0$ are two mutually perpendicular lines, these lines are the axes of the ellipse and their point of intersection is the centre of the ellipse. (You can also visualise it as shifting the origin and rotating the axes which will not affect the dimensions of the ellipse.)
Same can be observed in this question and we get the axes of the ellipse as $x+y=0$ and $x-y=0$ and the centre of the ellipse as $(0,0)$.
We can see that the given ellipse is geometrically identical (or congruent) to the ellipse $\frac{x^2}{9}+\frac{y^2}{1}=1$ (This ellipse obtained by evaluating the length of the axes of the given ellipse and then obtaining a similar ellipse centred at the origin with axes $x=0$ and $y=0$). Now since the question demands only a particular length in the answer, we can use this much simpler ellipse in standard form to solve the question.
Since the dimensions will remain unchanged, we have the coordinates of the points $A$ and $B$ as $A'(-2\sqrt2,0)$ and $B'(2\sqrt2,0)$ (Obtained by equating the distances of the points $A$ and $B$ from the centre of the original ellipse and then using them in the standard ellipse).
Now since you have the given coordinates and the ellipse in a much simpler form, can you solve the question easily without getting stuck in heavy calculations?