IIT-JEE 2010 · Paper 2 · Mathematics Q35

MathematicsSingle correct+3 / -1Question 35 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 orthocenter of the triangle PAB is

\left(5,\dfrac87\right)

\left(\dfrac75,\dfrac{25}8\right)

\left(\dfrac{11}5,\dfrac85\right)

\left(\dfrac8{25},\dfrac75\right)

Answer and worked solution

Official answer: C

Triangle $P(3,4)$ , $A(3,0)$ , $B\left(-\frac95,\frac85\right)$ .

Altitudes.

PA is vertical ( $x=3$ ), so the altitude from B is horizontal: $y=\frac85$ .
PB has slope $\dfrac{8/5-4}{-9/5-3}=\frac12$ , so the altitude from A has slope $-2$ : $y=-2(x-3)$ .

Intersection. $\frac85=-2(x-3)$ gives $x=\frac{11}5$ . Orthocentre $\left(\frac{11}5,\frac85\right)$ .

(C)

Topics:Equations of a line

IIT-JEE 2010 · Paper 2, Mathematics question 35