IIT-JEE 2010 · Paper 2 · Mathematics Q36

MathematicsSingle correct+3 / -1Question 36 of 57

Passage

Paragraph for Questions 34 to 36

Tangents are drawn from the point $P(3,4)$ to the ellipse $\dfrac{x^2}9+\dfrac{y^2}4=1$ touching the ellipse at points A and B.

The equation of the locus of the point whose distances from the point P and the line AB are equal, is

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

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

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

x^2+y^2-2xy+27x+31y-120=0

Answer and worked solution

Official answer: A

A parabola with focus $P(3,4)$ and directrix AB: $x+3y-3=0$ .
$(x-3)^2+(y-4)^2=\frac{(x+3y-3)^2}{10}.$

Expand. $10(x^2+y^2-6x-8y+25)=x^2+9y^2+6xy-6x-18y+9$ , which gives
$9x^2+y^2-6xy-54x-62y+241=0.$

(A)

Topics:Parabola Locus

IIT-JEE 2010 · Paper 2, Mathematics question 36