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JEE Mathematics Solved Question Paper 2016

Multiple Choice Questions

21.

The line AB cuts off equal intercepts 2a from the axes. From any point P on the line AB perpendiculars PR and PS are drawn on the axes. Locus of mid-point of RS is

x + y = a
x² + y² = 4a²
x² - y² = 2a²

22.

X + 8y - 22 = 0, 5x + 2y -34 = 0, 2x - 3y + 13= 0 are the three sides of a triangle. The area of the triangle is

36 sq units
19 sq units
42 sq units
72 sq units

23.

The line through the points (a, b) and (- a,- b) passes through the point

(1, 1)
(3a, - 2b)
(a², ab)
(a, b)

24.

The locus of the point of intersection of the straight lines where K is a non-zero real variable, is given by

a straight line
an ellipse
a parabola
a hyperbola

25.

The equation of a line parallel to the line 3x + 4y= 0 and touching the circle x² + y² = 9 in the first quadrant, is

3x +4y = 15
3x +4y = 45
3x +4y = 9
3x +4y = 27

26.

A line passing through the point of intersection of x + y = 4 and x - y = 2 makes an angle $\tan^{- 1} (\frac{3}{4})$ with the x-axis. It intersects the parabola y² = 4(x-3) at points $(x_{1}, y_{1}) and (x_{2}, y_{2})$ respectively. Then, $|x_{1} - x_{2}|$

$\frac{16}{9}$
$\frac{32}{9}$
$\frac{40}{9}$
$\frac{80}{9}$

27.

The equation of auxiliary circle of the ellipse 16x² + 25y² + 32x - 100y = 284 is

x² + y² + 2x - 4y - 20 = 0
x² + y² + 2x - 4y = 0
(x + 1)² + (y - 2)² = 400
(x + 1)² + (y - 2)² = 225

28.
If PQ is a double ordinate of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 such that ∆ OPQ$ is equilateral. O being the centre. Then, the eccentricity e satisfies

$1 < e < \frac{2}{\sqrt{3}}$

e = $\frac{2}{\sqrt{2}}$

e = $\frac{\sqrt{3}}{2}$

$e > \frac{2}{\sqrt{3}}$

$e > \frac{2}{\sqrt{3}}$

Given equation of hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and $∆$ OPQ is an equilateral triangle. PQ is double ordinate of the hyperbola.

Let the coordinates of P and Q be ( $asec (θ), btan (θ)$ ) and ( $asec (θ), - btan (θ)$ ), respectively.

$In ∆ OPQ, OP = PQ \Rightarrow a^{2} \sec^{2} (θ) + b^{2} \tan^{2} (θ) {(2 btan (θ))}^{2} \Rightarrow a^{2} \sec^{2} (θ) = 3 b^{2} \tan^{2} (θ) \Rightarrow \sin^{2} (θ) = \frac{a^{2}}{3 b^{2}}$

Now, $\sin^{2} (θ) < 1$

$\Rightarrow \frac{a^{2}}{3 b^{2}} < 1 \Rightarrow \frac{b^{2}}{a^{2}} > \frac{1}{3} \Rightarrow 1 + \frac{b^{2}}{a^{2}} > \frac{4}{3} ∴ e^{2} > \frac{4}{3} \Rightarrow e > \frac{2}{\sqrt{3}}$