JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Ellipse

The tangent and normal to the ellipse

3 x^{2} + 5 y^{2} = 32

at the point

P (2, 2)

meet the

x

-axis at

Q

and

R

, respectively. Then the area(in sq. units) of the triangle

P QR

Answer: (B)

Solution

Differentiating

3 x^{2} + 5 y^{2} = 32

gives

\frac{d y}{d x} = - \frac{3 x}{5 y}

, which at

P (2, 2)

- \frac{3}{5}

. The tangent

y - 2 = - \frac{3}{5} (x - 2)

meets the

x

-axis at

Q (\frac{16}{3}, 0)

, and the normal

y - 2 = \frac{5}{3} (x - 2)

meets it at

R (\frac{4}{5}, 0)

. Taking

QR

as the base and the

y

-coordinate of

P

as the height, the area is

\frac{1}{2} \frac{16}{3} - \frac{4}{5} \times 2 = \frac{68}{15}

square units.

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option