Question

$P$ lies on the line $y=x$ and $Q$ lies on $y=2x$. The equation for the locus of the midpoint of $PQ$, if $|PQ|=4$, is

• A. $25x^2+36xy+13y^2=4$
• B. $25x^2-36xy+13y^2=4$
• C. $25x^2-36xy-13y^2=4$
• D. $25x^2+36xy-13y^2=4$

Answer 1

The correct option is B $25x^2-36xy+13y^2=4$

P lies on $y=n$ So its coordinates $P(a,a)$

Q lies on $y=2n$ So its coordinates $Q(b,2b)$

given $|PQ|=4$

So $(a-b{)}^2+(a-2b{)}^2=4^2=\mathrm{16...}\left(1\right)$

Now Let M be the Mid Point of PQ

So coordinates of M $(\frac{a+b}{2},\frac{a+2b}{2})$

To trace M's Locus, we assign x and y coordinate

as! $n=\frac{a+b}{2}....\left(2\right)$

$y=\frac{a+2b}{2}....\left(3\right)$

Solving for a and b from $e{q}^n$ (2) and (3) we get

$b=2y-2n...\left(4\right)$

$a=4n-2y....\left(5\right)$

Substituting $e{q}^n$ (4) & (5) in $e{q}^n$ (1)

$16=(6n-4y{)}^2+(8n-6y{)}^2$

$4=(3n-2y{)}^2+(4n-3y{)}^2$

$25n^2-36ny+13y^2-4=0$

So option B is correct