ExamHabit
Practise free

JEE Main 2019 · 10 Apr · Shift II

JEE Main 2019 Mathematics question on Conic Sections

The tangent and normal to the ellipse 3x2+5y2=323x^2 + 5y^2 = 32 at the point P(2,2)P(2, 2) meets the X-axis at QQ and RR, respectively. Then, the area (in sq units) of the △PQR\triangle PQR is
  1. A.
    163\dfrac{16}{3}
  2. B.
    143\dfrac{14}{3}
  3. C.
    3415\dfrac{34}{15}
  4. D.
    6815\dfrac{68}{15}
Show answer and solution

Correct option: D

Solution

At P(2,2), differentiate 3x² + 5y² = 32 implicitly: 6x + 10y·y' = 0, so y' = -6x/(10y) = -6(2)/(10·2) = -3/5.

Tangent at P(2,2): y - 2 = (-3/5)(x - 2) → y = (-3/5)x + 6/5 + 2 = (-3/5)x + 16/5. At y=0: x = 16/3, so Q = (16/3, 0).

Normal at P(2,2) has slope 5/3: y - 2 = (5/3)(x - 2) → y = (5/3)x - 10/3 + 2 = (5/3)x - 4/3. At y=0: x = 4/5, so R = (4/5, 0).

Base QR = 16/3 - 4/5 = (80-12)/15 = 68/15. Height = y-coordinate of P = 2.

Area = (1/2)·(68/15)·2 = 68/15.

[JEE Main 2019 · 10 Apr · Shift II]

Practise Conic Sections the way it is asked

A timed set of real JEE Main Conic Sections questions, graded as the exam grades. No sign-up.

Start a Conic Sections test

More: all JEE Main Conic Sections questions · JEE Main 2019 · Mathematics chapter list