Conic Sections: Let E: (x^2/(16)+(y^2/…

Let E: (x^2/(16)+(y^2/(9)=1.

One or more options correct

Let

E:\ \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1

. Which of the following statements is/are correct?

A. The minimum area of the triangle bounded by a tangent to $E$ in the first quadrant and the coordinate axes is $12$
B. The two tangents to $E$ drawn from the point $(4,3)$ are perpendicular to each other
C. The length of the chord of contact of the tangents drawn from $(8,0)$ is $6$
D. The normal to $E$ at the point with eccentric angle $\pi/4$ meets the $x$ -axis at $\left(\dfrac{7\sqrt2}{4},0\right)$

Show answer and solution

Answer: A, B

(A) True: the tangent at

(4\cos\theta,3\sin\theta)

\frac{x\cos\theta}{4}+\frac{y\sin\theta}{3}=1

with intercepts

\frac4{\cos\theta}

and

\frac3{\sin\theta}

, so the area is

\frac{6}{\sin\theta\cos\theta}=\frac{12}{\sin2\theta}\ge12

, equality at

\theta=\frac\pi4

. (B) True: the director circle is

x^{2}+y^{2}=16+9=25

and

(4,3)

lies on it. (C) False: the chord of contact from

(8,0)

\frac{8x}{16}=1

, i.e.

x=2

, giving

y^{2}=9\left(1-\frac14\right)=\frac{27}4

and length

2\cdot\frac{3\sqrt3}{2}=3\sqrt3\ne6

. (D) False: the normal

\frac{4x}{\cos\theta}-\frac{3y}{\sin\theta}=7

meets

y=0

x=\frac74\cos\theta=\frac{7\sqrt2}{8}

, not

\frac{7\sqrt2}{4}

Original question written in the JEE Advanced 2016 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake