Question

Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $A$ and $B$.
The equation of the locus of the point whose distance from the point $P$ and the line $AB$ are equal, is?

• A. $9x^2+y^2-6xy-54x-62y+241=0$
• B. $x^2+9y^2+6xy-54x+62y-241=0$
• C. $9x^2+9y^2-6xy-54x-62y-241=0$
• D. $x^2+y^2-2xy+27x+31y-120=0$

Answer 1

Answer: (A)

$9x^2+y^2-6xy-54x-62y+241=0$

Chord of contact $AB,\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}-1=0$

$\frac{x}{3}+y-1=0$

$x=3(y-1)$

Let the point be $(h,k)$

Distance from the line

$\frac{|\frac{h}{3}+k-1|}{\sqrt{\frac{1}{9}+1}}$

$\frac{|h+3k-3|}{\sqrt{10}}$

Distance between $(3,4)$ and $(h,k)$

$⟹\sqrt{(h-3{)}^2+(k-4{)}^2}$

As they are equal,

$\frac{|h+3k-3|}{\sqrt{10}}=\sqrt{(h-3{)}^2+(k-4{)}^2}$

$\frac{(h+3k-3{)}^2}{10}=(h-3{)}^2+(k-4{)}^2$

$h^2+9k^2+9+6hk-6h-18k=10(h^2-6h+9+k^2-8k+16)$

$9h^2+k^2+241-54h-62k-6hk=0$

$9x^2+y^2-6xy+54x-62y+241=0$