Paragraph: Tangents are drawn from the point P(3,4) to the ellipse (x^2)/(9)+(y^2)/(4)=1 touching the ellipse at points A and B . Question: The orthocentre of the triangle P A B is

Correct answer: C. ((11)/(5), (8)/(5)) Equation of A B is aligned y-0 & =((8)/(5))/(-(9)/(5)-3)(x-3)=(8)/(-24)(x-3) y & =-(1)/(3)(x-3) x & +3 y=3 aligned Equation of the straight line perpendicular to A B through P is 3 x-y=5 . Equation of P A is x-3=0 . The equation of straight line perpendicular to P A through B((-9)/(5), (8)/(5)) . is y=(8)/(5) . Hence, the orthocentre is ((11)/(5), (8)/(5)) .

JEE Advanced 2010Paper 2

Exam: JEE Advanced
Year: 2010
Paper: Paper 2
Subject: Mathematics
Chapter: Straight Lines

JEE Advanced 2010 · Paper 2 · Mathematics · Straight Lines

Paragraph: Tangents are drawn from the point P(3,4) to the ellipse...

Paragraph: 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

.Question: The orthocentre of the triangle

P A B

Paragraph: Tangents are drawn from the point P(3,4) to the ellipse (x^2)/(9)+(y^2)/(4)=1 touching the ellipse at points

Options

A $\left(5, \frac{8}{7}\right)$
B $\left(\frac{7}{5}, \frac{25}{8}\right)$
C $\left(\frac{11}{5}, \frac{8}{5}\right)$
D $\left(\frac{8}{25}, \frac{7}{5}\right)$

Correct answer

C. $\left(\frac{11}{5}, \frac{8}{5}\right)$

Step-by-step solution

\text{ Equation of } A B \text{ is }

<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/vGNDL6PX8Om2ubizKilLEdDE0M3SMgzemqgZ4ITooP0.original.fullsize.png"><br>

\begin{aligned} y-0 & =\frac{\frac{8}{5}}{-\frac{9}{5}-3}(x-3)=\frac{8}{-24}(x-3) \\ \Rightarrow \quad y & =-\frac{1}{3}(x-3) \\ \Rightarrow \quad x & +3 y=3 \end{aligned}

Equation of the straight line perpendicular to A B

through

P

3 x-y=5

. Equation of

P A

x-3=0

. The equation of straight line perpendicular to

P A

through

B\left(\frac{-9}{5}, \frac{8}{5}\right)

. is

y=\frac{8}{5}

. Hence, the orthocentre is

\left(\frac{11}{5}, \frac{8}{5}\right)

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