If the tangent drawn to ellipse (x2/25)+(y2/16)=1 intersect the circle x2+y2=25 at A and B, then the locus of the point of intersection of tangents at A and B is

Question

If the tangent drawn to ellipse (x2/25)+(y2/16)=1 intersect the circle x2+y2=25 at A and B, then the locus of the point of intersection of tangents at A and B is (A) 25 x2+16 y2=225 (B) 25 x2+16 y2=625 (C) 16 x2+25 y2=225 (D) 16 x2+25 y2=625

Tardigrade Admin · Accepted Answer

Equation of AB: xh + yk =25....(1) Also, (x cos θ/5)+(y sin θ/4)=1.....(2) ∴ ( cos θ/5 h )=( sin θ/4 k )=(1/25) Use c2+s2=1 ⇒ (x2/25)+(y2/((25)4)2)=1

Q. If the tangent drawn to ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ intersect the circle $x^{2} + y^{2} = 25$ at $A$ and $B$ , then the locus of the point of intersection of tangents at $A$ and $B$ is

$25 x^{2} + 16 y^{2} = 225$

$25 x^{2} + 16 y^{2} = 625$

$16 x^{2} + 25 y^{2} = 225$

$16 x^{2} + 25 y^{2} = 625$

Equation of $A B : x h + y k = 25$ ....(1)
Also, $\frac{x c o s θ}{5} + \frac{y s i n θ}{4} = 1$ .....(2)
$∴ \frac{c o s θ}{5 h} = \frac{s i n θ}{4 k} = \frac{1}{25}$
Use $c^{2} + s^{2} = 1$
$\Rightarrow \frac{x ^{2}}{25} + \frac{y ^{2}}{( \frac{25}{4} ) ^{2}} = 1$

Q. If the tangent drawn to ellipse 25x2​+16y2​=1 intersect the circle x2+y2=25 at A and B, then the locus of the point of intersection of tangents at A and B is

25x2+16y2=225

25x2+16y2=625

16x2+25y2=225

16x2+25y2=625